Topic: Work, Energy, Power and Conservation Laws (45 Questions)
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Q.1 Q.1 If the values of force and length are increased four times then the unit of energy will increase by –
(1) 4 times
(2) 2 times
(3) 8 times
(4) 16 times
Correct Choice: Correct Answer: (4) 16 times
Step-by-Step Derivation: Energy = Force \times Displacement. If both Force and Length (displacement) are increased 4 times, Energy increases by 4 \times 4 = 16 times.
Q.2 A bullet of mass $P$ is fired with velocity $Q$ in a large body of mass $R$. The final velocity of the system will be:
(1) $\frac{P + R}{PQ}$
(2) $\frac{PQ}{P + R}$
(3) $\frac{PR}{P + Q}$
(4) $\frac{PQ}{R}$
Correct Choice: (2) $\frac{PQ}{P + R}$
Step-by-Step Derivation: By conservation of momentum, $P \times Q = (P + R) \times V$, hence the final velocity of the system $V = \frac{PQ}{P + R}$.
Q.3 A sphere of mass $m$ moving with a constant velocity collides with another stationary sphere of same mass. The ratio of velocities of two spheres after collision will be, if the co-efficient of restitution is $e$:
(1) $\frac{1 – e}{1 + e}$
(2) $\frac{1 + e}{1 – e}$
(3) $\frac{e}{1 + e}$
(4) $\frac{e}{1 – e}$
Correct Choice: (1) $\frac{1 – e}{1 + e}$
Step-by-Step Derivation: Using momentum conservation: $mu = m v_1 + m v_2 \implies v_1 + v_2 = u$. From the definition of coefficient of restitution: $e u = v_2 – v_1$. Solving for $v_1$ and $v_2$ yields $v_1 = \frac{u(1-e)}{2}$ and $v_2 = \frac{u(1+e)}{2}$. Thus, the ratio of their velocities after collision is $\frac{v_1}{v_2} = \frac{1 – e}{1 + e}$.
Q.4 An electric motor produces a tension of $4500\text{ N}$ in a load lifting cable and rolls it at the rate of $2\text{ m/s}$. The power of the motor is –
(1) $9\text{ kW}$
(2) $15\text{ kW}$
(3) $225\text{ kW}$
(4) $9 \times 10^3\text{ HP}$
Correct Choice: (1) $9\text{ kW}$
Step-by-Step Derivation: Power $P = F \times v = 4500\text{ N} \times 2\text{ m/s} = 9000\text{ W} = 9\text{ kW}$.
Q.5 A ball falls from a height of $5\text{ m}$ and strikes the roof of a lift. If at the time of collision, lift is moving in the upward direction with a velocity of $1\text{ m/s}$, then the velocity with which the ball rebounds after collision will be – ($e = 1$)
(1) $11\text{ m/s downwards}$
(2) $12\text{ m/s upwards}$
(3) $13\text{ m/s upwards}$
(4) $12\text{ m/s downwards}$
Correct Choice: (2) $12\text{ m/s upwards}$
Step-by-Step Derivation: Velocity of ball before collision $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 5} = 10\text{ m/s}$ (downwards). Relative velocity of approach $= 10 + 1 = 11\text{ m/s}$. Since $e = 1$, relative velocity of separation is also $11\text{ m/s}$. Velocity of ball after collision $v’ – 1 = 11 \implies v’ = 12\text{ m/s}$ (upwards).
Q.6 A force $\vec{F} = 5\hat{i} + 3\hat{j}\text{ N}$ is applied over a particle which displaces it from its origin to the point $\vec{r} = 2\hat{i} – \hat{j}\text{ m}$. The work done on the particle in joules is:
Q.7 Q.7 Two elastic bodies P and Q having equal masses are moving along the same line with velocities of 16 $m$/s and 10 $m$/s respectively. Their velocities after the elastic collision will be in $m$/s:
(1) 0 and 25
(2) 5 and 20
(3) 10 and 16
(4) 20 and 5
Correct Choice: Correct Answer: (3) 10 and 16
Step-by-Step Derivation: When two elastic bodies of equal masses undergo head-on elastic collision, they interchange their velocities. Therefore, the velocities after collision will be 10 m/s and 16 m/s.
Q.8 If the momentum of a body is increased $n$ times, its kinetic energy increases:
(1) $n\text{ times}$
(2) $2n\text{ times}$
(3) $\sqrt{n}\text{ times}$
(4) $n^2\text{ times}$
Correct Choice: (4) $n^2\text{ times}$
Step-by-Step Derivation: Kinetic Energy $E_K = \frac{p^2}{2m}$. If momentum $p$ is increased $n$ times ($p’ = np$), then $E_K’ = \frac{(np)^2}{2m} = n^2 \cdot E_K$. Thus, kinetic energy increases by $n^2$ times.
Q.9 Q.9 A metal ball does not rebound when struck on a wall, whereas a rubber ball of same mass when thrown with the same velocity on the wall rebounds. From this it is inferred that –
(1) Change in momentum is same in both
(2) Change in momentum in rubber ball is more
(3) Change in momentum in metal ball is more
(4) Initial momentum of metal ball is more than that of rubber ball.
Correct Choice: Correct Answer: (2) Change in momentum in rubber ball is more
Step-by-Step Derivation: For the metal ball, final velocity is 0, so change in momentum is m*v. For the rubber ball, it rebounds with some velocity (say, -v), so change in momentum is m*v – (-m*v) = 2m*v (which is larger).
Q.10 Q.10 The unit of the co – efficient of restitution is –
(1) m/s
(2) s/m
(3) m \times s
(4) None of the above
Correct Choice: Correct Answer: (4) None of the above
Step-by-Step Derivation: The coefficient of restitution is the ratio of relative velocity of separation to relative velocity of approach. It is a ratio of identical quantities and is therefore dimensionless and unitless.
Q.11 A bomb of mass $9\text{ kg}$ explodes into two pieces of $3\text{ kg}$ and $6\text{ kg}$. The velocity of $3\text{ kg}$ piece is $16\text{ m/s}$. The kinetic energy of $6\text{ kg}$ piece is –
(1) $768\text{ Joule}$
(2) $786\text{ Joule}$
(3) $192\text{ Joule}$
(4) $687\text{ Joule}$
Correct Choice: (3) $192\text{ Joule}$
Step-by-Step Derivation: By conservation of momentum: $m_1 v_1 = m_2 v_2 \implies 3 \times 16 = 6 \times v \implies v = 8\text{ m/s}$. Kinetic energy of $6\text{ kg}$ piece $= \frac{1}{2} m_2 v^2 = \frac{1}{2} \times 6 \times 8^2 = 3 \times 64 = 192\text{ J}$.
Q.12 Two solid rubber balls A and B whose masses are $200\text{ g}$ and $400\text{ g}$ respectively, are moving in mutually opposite directions. If the velocity of A is $0.3\text{ m/s}$ and both the balls come to rest after collision, then the velocity of ball B is –
Q.13 The graph between $\frac{p}{E_K}$ and $p$ is ($E_K$ = kinetic energy and $p$ = momentum) –
(1) Straight line
(2) Parabola
(3) Rectangular hyperbola
(4) Circle
Correct Choice: (3) Rectangular hyperbola
Step-by-Step Derivation: Since $E_K = \frac{p^2}{2m}$, we have $\frac{p}{E_K} = \frac{2m}{p}$. Let $y = \frac{p}{E_K}$ and $x = p$. Then $y = \frac{2m}{x} \implies y \cdot x = 2m = \text{constant}$. This is the equation of a rectangular hyperbola.
Q.14 A $1\text{ kg}$ ball falls from a height of $25\text{ cm}$ and rebounds up to a height of $9\text{ cm}$. The coefficient of restitution is:
Q.15 Q.15 The graph between potential energy $U$ and displacement $X$ in the state of stable equilibrium will be –
(1) A curve with a minimum (bowl-shaped opening upwards)
(2) A curve with a maximum (hill-shaped opening downwards)
(3) A straight line passing through origin
(4) A straight line parallel to displacement axis
Correct Choice: Correct Answer: (1) A curve with a minimum (bowl-shaped opening upwards)
Step-by-Step Derivation: In a state of stable equilibrium, the potential energy (U) is at a minimum value. Therefore, the graph of U vs X shows a curve with a minimum (bowl-shaped opening upwards).
Q.16 A force $F = (3x^2 + 2x – 7)\text{ N}$ acts on a $2\text{ kg}$ body as a result of which the body gets displaced from $x = 0$ to $x = 5\text{ m}$. The work done by the force will be –
Q.17 A $50\text{ g}$ bullet moving with a velocity of $10\text{ m/s}$ gets embedded into a $950\text{ g}$ stationary body. The loss in kinetic energy of the system will be –
(1) $5\%$
(2) $50\%$
(3) $100\%$
(4) $95\%$
Correct Choice: (4) $95\%$
Step-by-Step Derivation: Initial kinetic energy $K_1 = \frac{1}{2} m v^2 = \frac{1}{2} \times (0.05\text{ kg}) \times (10\text{ m/s})^2 = 2.5\text{ J}$. After embedding, by momentum conservation, final velocity $V = \frac{m v}{m+M} = \frac{50 \times 10}{50 + 950} = 0.5\text{ m/s}$. Final kinetic energy $K_2 = \frac{1}{2}(m+M)V^2 = \frac{1}{2} \times (1\text{ kg}) \times (0.5\text{ m/s})^2 = 0.125\text{ J}$. Percentage Loss in KE $= \frac{K_1 – K_2}{K_1} \times 100 = \frac{2.5 – 0.125}{2.5} \times 100 = 95\%$.
Q.18 Q.18 A crane lifts 300 kg weight from earth’s surface up to a height of 2 $m$ in 3 seconds. The average power generated by it will be –
(1) 1960 W
(2) 2205 W
(3) 4410 W
(4) 0 W
Correct Choice: Correct Answer: (1) 1960 W
Step-by-Step Derivation: Work Done = mgh = 300 \times 9.8 \times 2 = 5880 J. Power = Work / Time = 5880 / 3 = 1960 W.
Q.19 A body is dropped from a height $h$. When loss in its potential energy is $U$ then its velocity is $v$. The mass of the body is –
(1) $\frac{U^2}{2v}$
(2) $\frac{2v}{U}$
(3) $\frac{2v}{U^2}$
(4) $\frac{2U}{v^2}$
Correct Choice: (4) $\frac{2U}{v^2}$
Step-by-Step Derivation: By conservation of energy, Loss in PE = Gain in KE $\implies U = \frac{1}{2} m v^2 \implies m = \frac{2U}{v^2}$.
Q.20 Q.20 A block of mass 16 kg is moving on a frictionless horizontal surface with velocity 4 $m$/s and comes to rest after pressing a spring. If the force constant of the spring is 100 N/$m$ then the compression in the spring will be –
(1) 3.2 m
(2) 1.6 m
(3) 0.6 m
(4) 6.1 m
Correct Choice: Correct Answer: (2) 1.6 m
Step-by-Step Derivation: By conservation of energy: 1/2 \times m \times v^{2} = 1/2 \times k \times x^{2} => 16 \times 4^{2} = 100 \times x^{2} => 256 = 100 x^{2} => x^{2} = 2.56 => x = 1.6 m.
Q.21 A bomb initially at rest explodes by itself into three equal mass fragments. The velocities of two fragments are $(3\hat{i} + 2\hat{j})\text{ m/s}$ and $(-\hat{i} – 4\hat{j})\text{ m/s}$. Velocity of the third fragment is (in m/s):
(1) $2\hat{i} + 2\hat{j}$
(2) $2\hat{i} – 2\hat{j}$
(3) $-2\hat{i} + 2\hat{j}$
(4) $-2\hat{i} – 2\hat{j}$
Correct Choice: (3) $-2\hat{i} + 2\hat{j}$
Step-by-Step Derivation: By conservation of momentum, total momentum is conserved: $m\vec{v}_1 + m\vec{v}_2 + m\vec{v}_3 = 0 \implies \vec{v}_3 = -(\vec{v}_1 + \vec{v}_2) = -((3\hat{i} + 2\hat{j}) + (-\hat{i} – 4\hat{j})) = -(2\hat{i} – 2\hat{j}) = -2\hat{i} + 2\hat{j}\text{ m/s}$.
Q.22 Q.22 A ball of mass 1 kg is released from the tower of Pisa. The kinetic energy generated in it after falling through 10 $m$ will be –
(1) 10 J
(2) 9.8 J
(3) 0.98 J
(4) 98 J
Correct Choice: Correct Answer: (4) 98 J
Step-by-Step Derivation: Kinetic Energy generated = Potential Energy lost = mgh = 1 \times 9.8 \times 10 = 98 J.
Q.23 Q.23 A 10 kg satellite completes one revolution around the earth at a height of 100 km in 108 minutes. The work done by the gravitational force of earth will be –
(1) 108 \times 100 \times 10 J
(2) 10^{8} J
(3) 0 J
(4) 10^{4} J
Correct Choice: Correct Answer: (3) 0 J
Step-by-Step Derivation: The gravitational force on the satellite acts as a centripetal force which is always perpendicular to the direction of motion (displacement). Hence, the work done is zero (W = F \cdot ds = F ds cos(90 ^\circ ) = 0).
Q.24 A particle moves in a potential region given by $U = 8x^2 – 4x + 400\text{ J}$. Its state of equilibrium will be –
Q.25 A person of mass $m$ is standing on one end of a plank of mass $M$ and length $L$ and floating in water. The person moves from one end to another and stops. The displacement of the plank is –
(1) $\frac{mL}{M + m}$
(2) $\frac{ML}{M + m}$
(3) $\frac{mL}{M}$
(4) $\frac{ML}{m}$
Correct Choice: (1) $\frac{mL}{M + m}$
Step-by-Step Derivation: Since there is no external horizontal force on the system, the center of mass remains stationary. If the person moves by $L$ relative to the plank, and the plank shifts by $x$ relative to the water, the net center of mass displacement is zero: $m(L – x) = M x \implies mL = (M + m)x \implies x = \frac{mL}{M + m}$.
Q.26 A bullet of mass $m$ moving with a speed $v$ strikes a wooden block of mass $M$ and gets embedded into the block. The final speed is:
(1) $\frac{M v}{M + m}$
(2) $\frac{(M + m)v}{m}$
(3) $\frac{m v}{M + m}$
(4) $\frac{m v}{M}$
Correct Choice: (3) $\frac{m v}{M + m}$
Step-by-Step Derivation: By conservation of linear momentum: $m \cdot v = (M + m) \cdot U \implies \text{final speed } U = \frac{mv}{M+m}$.
Q.27 Two men with weights in the ratio $5 : 3$ run up a staircase in times in the ratio $11 : 9$. The ratio of power of first to that of second is –
(1) $15 / 11$
(2) $11 / 15$
(3) $11 / 9$
(4) $9 / 11$
Correct Choice: (1) $15 / 11$
Step-by-Step Derivation: Power $P = \frac{W}{t} = \frac{mgh}{t}$. The ratio of their powers is $\frac{P_1}{P_2} = \left(\frac{w_1}{w_2}\right) \times \left(\frac{t_2}{t_1}
ight) = \left(\frac{5}{3}\right) \times \left(\frac{9}{11}\right) = \frac{15}{11}$.
Q.28 Q.28 The retarding force required to reduce velocity of a 3 kg body from 0.75 $m$/s to 0.25 $m$/s in 0.02 sec will be –
(1) 25 N
(2) 50 N
(3) 75 N
(4) 100 N
Correct Choice: Correct Answer: (3) 75 N
Step-by-Step Derivation: Acceleration a = (v – u) / t = (0.25 – 0.75) / 0.02 = -0.50 / 0.02 = -25 m/s^{2}. Retarding force = m \times |a| = 3 kg \times 25 m/s^{2} = 75 N.
Q.29 Q.29 A 2 kg mass lying on a table is displaced in the horizontal direction through 50 cm. The work done by the normal reaction will be –
(1) 0
(2) 100 Joule
(3) 100 erg
(4) 10 Joule
Correct Choice: Correct Answer: (1) 0
Step-by-Step Derivation: The normal reaction force is directed vertically upwards, while the displacement is purely horizontal. Since the angle between them is 90 ^\circ , Work Done = F ds cos(90 ^\circ ) = 0.
Q.30 Q.30 A car is moving with a speed of 40 km/hr. If the car engine generates 7 kilowatt power, then the resistance in the path of motion of the car will be –
(1) 360 Newton
(2) 630 Newton
(3) Zero
(4) 280 Newton
Correct Choice: Correct Answer: (2) 630 Newton
Step-by-Step Derivation: Velocity v = 40 km/hr = 40 \times (5/18) = 100/9 m/s. Power P = F \times v => 7000 W = F \times (100/9) => F = 70 \times 9 = 630 N.
Q.31 Q.31 Figure shows the vertical section of a frictionless surface. A block of mass 2 kg is released from the position A (height = 10 $m$); its kinetic energy as it reaches the position C (height = 3 $m$) is:
(1) 180 J
(2) 140 J
(3) 40 J
(4) 280 J
Correct Choice: Correct Answer: (2) 140 J
Step-by-Step Derivation: By conservation of mechanical energy: Loss in PE = Gain in KE => KE = mg(h_A – h_C) = 2 \times 10 \times (10 – 3) = 2 \times 10 \times 7 = 140 J.
Q.32 A force $F = Kx^2$ acts on a particle at an angle of $60^\circ$ with the $x$-axis. The work done in displacing the particle from $x_1$ to $x_2$ will be –
(1) $\frac{1}{2} K (x_2^3 – x_1^3)$
(2) $\frac{1}{3} K (x_2^3 – x_1^3)$
(3) $\frac{1}{6} K (x_2^3 – x_1^3)$
(4) $\frac{1}{4} K (x_2^2 – x_1^2)$
Correct Choice: (3) $\frac{1}{6} K (x_2^3 – x_1^3)$
Step-by-Step Derivation: $dW = F \, dx \cos(60^\circ) = Kx^2 \, dx \times 0.5$. Integrating from $x_1$ to $x_2$: $W = \int_{x_1}^{x_2} 0.5 K x^2 \, dx = 0.5 K \left[ \frac{x^3}{3} \right]_{x_1}^{x_2} = \frac{1}{6} K (x_2^3 – x_1^3)$.
Q.33 A ball moving with velocity of $9\text{ m/s}$ collides with another similar stationary ball. After the collision both the balls move in directions making an angle of $30^\circ$ with the initial direction. After the collision their speed will be:
(1) $2.6\text{ m/s}$
(2) $5.2\text{ m/s}$
(3) $0.52\text{ m/s}$
(4) $52\text{ m/s}$
Correct Choice: (2) $5.2\text{ m/s}$
Step-by-Step Derivation: By conservation of momentum along the initial direction: $m u = m v_1 \cos(30^\circ) + m v_2 \cos(30^\circ)$. Since the balls are identical and move symmetrically, their post-collision speeds are equal ($v_1 = v_2 = v$). Thus: $u = 2v \cos(30^\circ) \implies 9 = 2v \left(\frac{\sqrt{3}}{2}
ight) = v\sqrt{3} \implies v = \frac{9}{\sqrt{3}} = 3\sqrt{3} \approx 5.2\text{ m/s}$.
Q.34 A bomb of $50\text{ kg}$ is fired from a cannon with a velocity $600\text{ m/s}$. If the mass of the cannon is $10^3\text{ kg}$, then its recoil velocity will be –
Q.35 Two masses $m_1 = 2\text{ kg}$ and $m_2 = 5\text{ kg}$ are moving on a frictionless surface with velocities $10\text{ m/s}$ and $3\text{ m/s}$ respectively. $m_2$ is ahead of $m_1$. An ideal spring of spring constant $k = 1120\text{ N/m}$ is attached on the back side of $m_2$. The maximum compression of the spring will be:
(1) $0.51\text{ m}$
(2) $0.062\text{ m}$
(3) $0.25\text{ m}$
(4) $0.72\text{ m}$
Correct Choice: (3) $0.25\text{ m}$
Step-by-Step Derivation: Maximum compression happens when both masses have the same common velocity $V$. By momentum conservation: $V = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} = \frac{2(10) + 5(3)}{2+5} = \frac{35}{7} = 5\text{ m/s}$. By conservation of energy: $\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}(m_1+m_2)V^2 + \frac{1}{2}k x^2 \implies \frac{1}{2}(2)(10^2) + \frac{1}{2}(5)(3^2) = \frac{1}{2}(7)(5^2) + \frac{1}{2}(1120)x^2 \implies 100 + 22.5 = 87.5 + 560x^2 \implies 35 = 560x^2 \implies x^2 = \frac{35}{560} = \frac{1}{16} \implies x = 0.25\text{ m}$.
Q.36 A solid sphere is moving and it makes an elastic collision with another stationary sphere of half of its own radius. After collision it comes to rest. The ratio of the densities of materials of second sphere and first sphere is –
(1) $2$
(2) $4$
(3) $8$
(4) $16$
Correct Choice: (3) $8$
Step-by-Step Derivation: Since the first sphere comes to rest after elastic head-on collision with a stationary sphere, their masses must be identical: $m_1 = m_2$. Mass is density times volume: $\rho_1 \left( \frac{4}{3}\pi r_1^3 \right) = \rho_2 \left( \frac{4}{3}\pi r_2^3 \right) \implies \frac{\rho_2}{\rho_1} = \left( \frac{r_1}{r_2} \right)^3$. Given $r_2 = \frac{r_1}{2} \implies \frac{r_1}{r_2} = 2$, therefore $\frac{\rho_2}{\rho_1} = (2)^3 = 8$.
Q.37 Q.37 The mass of a bucket full of water is 15 kg. It is being pulled up from a 15 $m$ deep well. Due to a hole in the bucket 6 kg water flows out of the bucket. The work done in drawing the bucket out of the well will be –
(1) 900 Joule
(2) 1500 Joule
(3) 1800 Joule
(4) 2100 Joule
Correct Choice: Correct Answer: (3) 1800 Joule
Step-by-Step Derivation: Initial mass is 15 kg, final mass is 15 – 6 = 9 kg. Since water leaks at a uniform rate, the average mass pulled is (15 + 9)/2 = 12 kg. Work Done = m_avg \times g \times h = 12 kg \times 10 m/s^{2} \times 15 m = 1800 J.
Q.38 Q.38 A 5 kg body collides with another stationary body. After the collision, the bodies move in the same direction with one – third of the velocity of the first body. The mass of the second body will be –
(1) 5 kg
(2) 10 kg
(3) 15 kg
(4) 20 kg
Correct Choice: Correct Answer: (2) 10 kg
Step-by-Step Derivation: Initial momentum = m1 \times u1 = 5 \times u1. After collision, both move with velocity u1/3 in the same direction. By conservation of momentum: 5 \times u1 = (5 + m2) \times (u1/3) => 15 = 5 + m2 => m2 = 10 kg.
Q.39 A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a_c$ is varying with time $t$ as $a_c = k^2 r t^2$, where $k$ is a constant. The power delivered to the particle by the forces acting on it will be –
(1) $m k^2 t^2 r$
(2) $m k^2 r^2 t^2$
(3) $\text{zero}$
(4) $m k^2 r^2 t$
Correct Choice: (4) $m k^2 r^2 t$
Step-by-Step Derivation: Centripetal acceleration $a_c = \frac{v^2}{r} = k^2 r t^2 \implies v = k r t$. Tangential acceleration $a_t = \frac{dv}{dt} = k r$. The tangential force causing this acceleration is $F_t = m a_t = m k r$. Power delivered to the particle is $P = F_t \cdot v = (m k r) \cdot (k r t) = m k^2 r^2 t$.
Q.40 Q.40 A 10 g bullet, moving with a velocity of 500 $m$/s, enters a stationary piece of ice of mass 10 kg and stops. If the piece of ice is lying on a frictionless plane, then its velocity will be:
(1) 5 cm/s
(2) 5 m/s
(3) 0.5 m/s
(4) 0.5 cm/s
Correct Choice: Correct Answer: (3) 0.5 m/s
Step-by-Step Derivation: By conservation of momentum: m*u = (M + m)*v => (0.01 kg) \times 500 m/s = (10 + 0.01) kg \times v => 5 \approx 10*v => v \approx 0.5 m/s.
Q.41 Q.41 A 5 gm lump of clay, moving with a velocity of 10 cm/s towards east, collides head – on with another 2 gm lump of clay moving with 15 cm/s towards west. After collision, the two lumps stick together. The velocity of the compound lump will be –
(1) 5 cm/s towards east
(2) 5 cm/s towards west
(3) 2.88 cm/s towards east
(4) 2.5 cm/s towards west
Correct Choice: Correct Answer: (3) 2.88 cm/s towards east
Step-by-Step Derivation: Let East be positive direction. Initial momentum = 5 \times 10 + 2 \times (-15) = 50 – 30 = +20 gm cm/s. Combined mass = 5 + 2 = 7 gm. Final velocity v = 20 / 7 \approx 2.86 cm/s (option 2.88 cm/s) towards east.
Q.42 A frictionless steel ball of radius $2\text{ cm}$, moving on a horizontal plane with a velocity of $5\text{ cm/s}$, collides head–on with another stationary steel ball of radius $3\text{ cm}$. The velocities of two bodies after collision will respectively be (in cm/s) ($e = 1$) –
(1) $2.7$, $2.3$
(2) $-2.7$, $2.3$
(3) $2.7$, $-2.3$
(4) $-2.7$, $-2.3$
Correct Choice: (2) $-2.7$, $2.3$
Step-by-Step Derivation: Mass is proportional to volume, so $m \propto r^3$. Therefore, $m_1 \propto 2^3 = 8$ and $m_2 \propto 3^3 = 27$. Let $m_1 = 8k$ and $m_2 = 27k$. For a head-on elastic collision with a stationary target: $v_1 = \frac{m_1 – m_2}{m_1 + m_2} u_1 = \frac{8 – 27}{35} \times 5 = -\frac{19}{7} \approx -2.7\text{ cm/s}$. $v_2 = \frac{2 m_1}{m_1 + m_2} u_1 = \frac{16}{35} \times 5 = \frac{16}{7} \approx 2.3\text{ cm/s}$.
Q.43 A force acts on a $30\text{ g}$ particle in such a way that the position of the particle as a function of time is given by $x = 3t – 4t^2 + t^3$, where $x$ is in metres and $t$ is in seconds. The work done during the first 4 seconds is:
Q.44 A man is supplying an instantaneous power of $500\text{ J/s}$ to a massless string by pulling it at an instantaneous speed of $10\text{ m/s}$. It is known that kinetic energy of the block is increasing at a rate of $100\text{ J/s}$ at that instant. Then the mass of the block is –
(1) $5\text{ kg}$
(2) $3\text{ kg}$
(3) $10\text{ kg}$
(4) $4\text{ kg}$
Correct Choice: (4) $4\text{ kg}$
Step-by-Step Derivation: Power supplied to string $P = F \cdot v \implies 500 = F \times 10 \implies \text{Tension } T = F = 50\text{ N}$. The rate of increase of KE of the block is given by $P_{\text{block}} = F_{\text{net}} \cdot v \implies (T – mg)v = 100 \implies (50 – m \times 10) \times 10 = 100 \implies 50 – 10m = 10 \implies 10m = 40 \implies m = 4\text{ kg}$.
Q.45 A shell is fired from a cannon with velocity $V\text{ m/s}$ at an angle $\theta$ with the horizontal direction. At the highest point in its path with same speed it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon. The speed in m/sec. of the other piece immediately after the explosion is:
(1) $V \cos \theta$
(2) $3V \cos \theta$
(3) $2V \cos \theta$
(4) $\frac{3}{2} V \cos \theta$
Correct Choice: (2) $3V \cos \theta$
Step-by-Step Derivation: At the highest point, the velocity is purely horizontal: $v_x = V \cos \theta$. Initial momentum of the shell (mass $2m$) $= 2m(V \cos \theta)$. At explosion, one piece (mass $m$) retraces its path, meaning its velocity becomes $-V \cos \theta$. Let the velocity of the second piece be $V’$. By conservation of linear momentum: $2m(V \cos \theta) = m(-V \cos \theta) + m V’ \implies 2V \cos \theta = -V \cos \theta + V’ \implies V’ = 3V \cos \theta$.
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