JEE Physics Collisions MCQs Set 03

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Chapter-wise Objective Questions: Collisions

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Multiple Choice Questions

Question 1: Choose the correct statement from the following:
(i) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(ii) The work done by a force on a body in nature, over a closed loop, is always zero.
(iii) In an inelastic collision of two bodies, the final kinetic energy is less than the initial kinetic energy of the system.
(a) (i)
(b) (ii)
(c) (iii)
(d) all
Show Answer & Explanation

Answer: (c) (iii)

Explanation:
1. In an elastic collision the momentum and energy of the system are conserved, not of each body.
2. The work over a closed loop is zero only for conservative forces.
3. Kinetic energy is lost in an inelastic collision, so (iii) is true.

Teacher's Note:
1. Conservation laws apply to the whole isolated system.
2. Friction does non-zero work over a closed loop.

Question 2: Consider two statements: (A) The linear momentum of a system of particles is zero. (B) The kinetic energy of a system of particles is zero. Then:
(a) A implies B but B does not imply A
(b) A does not imply B but B implies A
(c) A implies B and B implies A
(d) A does not imply B and B does not imply A
Show Answer & Explanation

Answer: (b) A does not imply B but B implies A

Explanation:
1. Zero kinetic energy means every particle is at rest, so the momentum is zero: B implies A.
2. Zero momentum is possible with moving particles (for example, equal and opposite momenta), so A does not imply B.

Teacher's Note:
1. Kinetic energy is a sum of non-negative terms.
2. Momentum is a vector sum and can cancel.

Question 3: A massive ball moving with speed v collides head-on with a tiny ball of mass much smaller than the first ball, which is at rest. If the collision is completely elastic, the speed of the second ball after the collision is approximately:
(a) v
(b) 2v
(c) \( \infty \)
(d) \( v/2 \)
Show Answer & Explanation

Answer: (b) 2v

Explanation:
1. For m1 ≫ m2 with m2 at rest, v2 = 2m1v/(m1 + m2) ≈ 2v.

Teacher's Note:
1. The heavy ball is almost unaffected.
2. Relative speed of approach equals relative speed of separation.

Question 4: An object of mass 3m splits into three equal fragments. Two fragments have velocities \( v\hat j \) and \( v\hat i \). The velocity of the third fragment is:
(a) \( v(\hat j-\hat i) \)
(b) \( -v(\hat i+\hat j) \)
(c) \( v(\hat i-\hat j) \)
(d) \( \frac{v(\hat i+\hat j)}{\sqrt2} \)
Show Answer & Explanation

Answer: (b) \( -v(\hat i+\hat j) \)

Explanation:
1. The initial momentum is zero (the object is at rest), so the total final momentum is zero.
2. m(v ĵ) + m(v î) + m v3 = 0, so v3 = -v(î + ĵ).

Teacher's Note:
1. Momentum is conserved in a splitting of an isolated body.
2. The three fragments have the same mass.

Question 5: Two particles have position vectors \( \vec r_1=(3\hat i+5\hat j) \) m and \( \vec r_2=(-5\hat i-3\hat j) \) m and are moving with velocities \( \vec v_1=(4\hat i+3\hat j) \) m/s and \( \vec v_2=(a\hat i+7\hat j) \) m/s. If they collide after 2 seconds, the value of a is:
(a) 4
(b) 6
(c) 8
(d) 2
Show Answer & Explanation

Answer: (c) 8

Explanation:
1. They meet when r1 + 2v1 = r2 + 2v2.
2. x: 3 + 8 = -5 + 2a, so a = 8. (y: 5 + 6 = -3 + 14 = 11, which checks.)

Teacher's Note:
1. Equal position vectors at the instant of collision.
2. The y-components confirm the time.

Question 6: A bomb of mass 12 kg explodes into two pieces of masses 4 kg and 8 kg. The velocity of the 8 kg mass is 6 m/s. The kinetic energy of the other mass is:
(a) 288 J
(b) 32 J
(c) 48 J
(d) 24 J
Show Answer & Explanation

Answer: (a) 288 J

Explanation:
1. Momentum: 4v = 8 × 6, so v = 12 m/s.
2. KE = (1/2)(4)(12)² = 288 J.

Teacher's Note:
1. The bomb is at rest initially, so the momenta are equal and opposite.
2. Compute the kinetic energy of the lighter piece.

Question 7: A ball A of mass 1 kg, moving with a speed of 12 m/s, collides obliquely and elastically with another ball B which was initially at rest. Ball A then moves off at right angles to its original direction with a speed of 5 m/s. The magnitude of the momentum of ball B after the collision is:
(a) 5 kg m/s
(b) 13 kg m/s
(c) 17 kg m/s
(d) 11 kg m/s
Show Answer & Explanation

Answer: (b) 13 kg m/s

Explanation:
1. Initial momentum 12 (along x). After: A has 5 (along y).
2. B's momentum = (12, -5), magnitude = √(144 + 25) = 13 kg m/s.

Teacher's Note:
1. Momentum is conserved as a vector.
2. Use Pythagoras on the components.

Question 8: A player catches a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to:
(a) 30 N
(b) 300 N
(c) 150 N
(d) 3 N
Show Answer & Explanation

Answer: (a) 30 N

Explanation:
1. F = Δp/Δt = 0.15 × 20/0.1 = 30 N.

Teacher's Note:
1. Force equals the rate of change of momentum.
2. Convert grams to kilograms.

Question 9: A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments, each of mass m, are found to move with a speed v each in mutually perpendicular directions. The total energy released in the process is:
(a) \( \frac32mv^2 \)
(b) \( \frac52mv^2 \)
(c) \( mv^2 \)
(d) \( \frac12mv^2 \)
Show Answer & Explanation

Answer: (a) \( \frac32mv^2 \)

Explanation:
1. Third fragment has mass 2m and momentum √2 mv, so its speed is v/√2.
2. Energy = 2(1/2)mv² + (1/2)(2m)(v²/2) = mv² + (1/2)mv² = (3/2)mv².

Teacher's Note:
1. Find the third fragment's velocity from momentum.
2. Add the three kinetic energies.

Question 10: A shell of mass 200 g is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is:
(a) 80 m/s
(b) 100 m/s
(c) 40 m/s
(d) 120 m/s
Show Answer & Explanation

Answer: (b) 100 m/s

Explanation:
1. Momentum: 0.2v = 4V, so V = 0.05v.
2. Energy: (1/2)(0.2)v² + (1/2)(4)(0.05v)² = 0.105v² = 1050, so v = 100 m/s.

Teacher's Note:
1. Momentum conservation gives the recoil speed.
2. Total energy given equals the sum of kinetic energies.

Question 11: A body of mass 10 kg is released from rest at a height of 10 m on a smooth curved surface. It then moves along a horizontal surface 10 m long with coefficient of friction 0.1 and strikes a spring of force constant k = 2 × 10⁴ N/m fixed at the end of this surface. The compression of the spring is approximately (g = 10 m/s²):
(a) 1 m
(b) 0.2 m
(c) 0.3 m
(d) 0.1 m
Show Answer & Explanation

Answer: (c) 0.3 m

Explanation:
1. Energy at the end of the rough part = mgh - μmgd = 1000 - 100 = 900 J.
2. (1/2)kx² = 900, so x² = 1800/(2 × 10⁴) = 0.09 and x = 0.3 m.

Teacher's Note:
1. Friction removes energy over the 10 m rough path.
2. Equate the remaining kinetic energy to the spring energy.

Question 12: A \( {}^{238}\mathrm{U} \) nucleus decays by emitting an alpha particle of speed v m/s. The recoil speed of the residual nucleus (in m/s) is:
(a) \( -4v/238 \)
(b) \( v/4 \)
(c) \( -4v/234 \)
(d) \( 4v/238 \)
Show Answer & Explanation

Answer: (c) \( -4v/234 \)

Explanation:
1. The residual nucleus has mass number 234.
2. Momentum conservation: 4v + 234V = 0, so V = -4v/234 (opposite to the alpha).

Teacher's Note:
1. Mass number falls by 4 in alpha decay.
2. Momenta are equal and opposite.

Question 13: A particle of mass m moving horizontally with speed 6 m/s collides elastically (head-on, in one dimension) with a much heavier particle M (m ≪ M) which is moving in the same direction at 4 m/s. The speed of the lighter particle after the collision will be:
(a) 2 m/s in the original direction
(b) 14 m/s opposite to the original direction
(c) 2 m/s opposite to the original direction
(d) 4 m/s in the original direction
Show Answer & Explanation

Answer: (a) 2 m/s in the original direction

Explanation:
1. For m ≪ M the heavy particle's speed is unchanged, and the relative velocity reverses.
2. Relative velocity before = 6 - 4 = 2 m/s, after = -2 m/s, so v = 4 - 2 = 2 m/s in the original direction.

Teacher's Note:
1. In the frame of the heavy particle the light one bounces back.
2. Transform back to the ground frame.

Question 14: Two identical masses M moving with velocities u1 and u2 collide perfectly inelastically. The loss in energy is:
(a) \( \frac M2(u_1-u_2)^2 \)
(b) \( \frac M2(u_2-u_1)^2 \)
(c) \( \frac M4(u_1-u_2)^2 \)
(d) \( \frac M8(u_1-u_2)^2 \)
Show Answer & Explanation

Answer: (c) \( \frac M4(u_1-u_2)^2 \)

Explanation:
1. Common velocity = (u1 + u2)/2.
2. Loss = (1/2)M(u1² + u2²) - (1/2)(2M)((u1 + u2)/2)² = (M/4)(u1 - u2)².

Teacher's Note:
1. Use the common velocity after the collision.
2. The loss depends on the relative velocity only.

Question 15: In a gamma decay process, the internal energy of a nucleus of mass M decreases, and a gamma photon of energy E and linear momentum E/c is emitted and the nucleus recoils. The decrease in internal energy is:
(a) zero
(b) \( E+\frac{E^2}{2Mc^2} \)
(c) \( \frac{E^2}{2Mc^2} \)
(d) E
Show Answer & Explanation

Answer: (b) \( E+\frac{E^2}{2Mc^2} \)

Explanation:
1. The nucleus recoils with momentum E/c, so its kinetic energy is (E/c)²/(2M) = E²/(2Mc²).
2. Decrease in internal energy = E + E²/(2Mc²).

Teacher's Note:
1. Energy released goes to the photon and to the recoil.
2. Use KE = p²/2M for the nucleus.

Question 16: A trolley of mass 200 kg moves with a uniform speed of 10 m/s on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other with a speed of 4 m/s relative to the trolley in a direction opposite to its motion, and jumps out of the trolley. The final speed of the trolley is:
(a) 10.4 m/s
(b) 12.4 m/s
(c) 16 m/s
(d) none of these
Show Answer & Explanation

Answer: (a) 10.4 m/s

Explanation:
1. Child's velocity after jumping (ground frame) = V - 4, where V is the trolley's final speed.
2. Momentum: 220(10) = 200V + 20(V - 4), so 220V = 2280 and V = 10.36 m/s ≈ 10.4 m/s.

Teacher's Note:
1. Convert the relative velocity to the ground frame.
2. Momentum is conserved horizontally.

Question 17: A car of mass 200 kg is rolling at a speed of 1 m/s towards a spring-stop system. The spring is non-linear such that it develops a force of 300x² N for a deflection of x metres. The maximum deceleration that the car undergoes is:
(a) 2.5 m/s²
(b) 1.5 m/s²
(c) 1 m/s²
(d) 2 m/s²
Show Answer & Explanation

Answer: (b) 1.5 m/s²

Explanation:
1. Kinetic energy = (1/2)(200)(1)² = 100 J = ∫300x² dx = 100x³, so x = 1 m.
2. Maximum force = 300 × 1 = 300 N.
3. Maximum deceleration = 300/200 = 1.5 m/s².

Teacher's Note:
1. Equate the work done by the spring to the initial kinetic energy.
2. The deceleration is largest at the maximum compression.

Question 18: A ball of mass 1 kg bounces against the ground. It approaches the horizontal ground with a speed of 25 m/s at 45° to the ground and leaves with a speed of 20 m/s at 30° to the ground (the horizontal motion continues in the same sense). The impulse exerted on the ball is approximately:
(a) 27.68 N s
(b) 31 N s
(c) 7.8 N s
(d) 62 N s
Show Answer & Explanation

Answer: (a) 27.68 N s

Explanation:
1. Before: vx = 25 cos 45° = 17.68, vy = -17.68 m/s. After: vx = 20 cos 30° = 17.32, vy = +10 m/s.
2. Δp = (-0.36, 27.68), magnitude ≈ 27.68 N s.

Teacher's Note:
1. Impulse equals the change in momentum (a vector).
2. Resolve the velocities before and after.

Question 19: An intense stream of water of cross-sectional area A strikes a wall at an angle θ with the normal to the wall and returns back elastically. If the density of water is ρ and its velocity is v, then the force exerted on the wall will be:
(a) \( 2Av^2\rho \)
(b) \( 2Av^2\rho\cos^2\theta \)
(c) \( 2Av^2\rho\cos\theta \)
(d) \( 2Av\rho \)
Show Answer & Explanation

Answer: (c) \( 2Av^2\rho\cos\theta \)

Explanation:
1. Mass of water striking per second = ρAv.
2. The normal component of velocity reverses: Δv = 2v cos θ.
3. F = ρAv × 2v cos θ = 2Av²ρ cos θ.

Teacher's Note:
1. Only the component of velocity along the normal changes.
2. Force = (mass per second) × (change in velocity).

Question 20: A particle of mass 2m is projected at an angle of 45° with the horizontal with a velocity of \( 20\sqrt2 \) m/s. After 1 second the explosion takes place and the particle is broken into two equal pieces. As a result of the explosion one part comes to rest. The maximum height from the ground attained by the other part is (g = 10 m/s²):
(a) 50 m
(b) 35 m
(c) 40 m
(d) 25 m
Show Answer & Explanation

Answer: (b) 35 m

Explanation:
1. Components are (20, 20) m/s. At t = 1 s: velocity (20, 10) and height = 20 - 5 = 15 m.
2. Momentum: 2m(20, 10) = m v', so v' = (40, 20) m/s.
3. Further rise = 20²/(2 × 10) = 20 m, so the maximum height = 15 + 20 = 35 m.

Teacher's Note:
1. Momentum is conserved in the explosion.
2. The vertical component of the second piece's velocity decides the extra height.

Question 21: Two balls having masses m and 2m are fastened to two light strings of the same length l. The other ends of the strings are fixed at the same point O. The strings are held horizontal on opposite sides of O and the system is released from rest. The collision between the balls (at the lowest point) is elastic. The speed of the ball of mass 2m just after the collision is:
(a) \( \frac{\sqrt{2gl}}{3} \)
(b) \( \frac{\sqrt{50gl}}{3} \)
(c) \( 2\sqrt{\frac{2gl}{3}} \)
(d) \( \frac{2\sqrt{50gl}}{3} \)
Show Answer & Explanation

Answer: (a) \( \frac{\sqrt{2gl}}{3} \)

Explanation:
1. Each ball reaches the lowest point with speed u = √(2gl), moving in opposite directions.
2. Elastic collision: v(2m) = [(2m - m)(-u) + 2m(u)]/(3m) = u/3.
3. So the speed of the 2m ball is √(2gl)/3.

Teacher's Note:
1. Both balls fall through the same height, so they have the same speed at the bottom.
2. Use the elastic collision formula for the second ball.

Question 22: A bomb at rest explodes into a large number of tiny fragments. Then:
(i) the momentum of all the fragments is zero
(ii) the momentum of all the fragments increases
(iii) the kinetic energy of all the fragments is zero
(iv) the kinetic energy of all the fragments is more than zero
Which of the following is/are correct?
(a) i, iv
(b) iii, iv
(c) ii, iii
(d) i, ii
Show Answer & Explanation

Answer: (a) i, iv

Explanation:
1. The total momentum stays zero.
2. The total kinetic energy increases from zero to a positive value.

Teacher's Note:
1. Explosions conserve momentum but not kinetic energy.
2. The internal chemical energy becomes kinetic.

Question 23: In an elastic head-on collision of two identical bodies:
(i) the momenta are interchanged
(ii) the velocities are interchanged
(iii) the kinetic energies are interchanged
(iv) none of the above
Which of the following is/are correct?
(a) i, ii, iii
(b) ii, iii, iv
(c) i, iii, iv
(d) i, ii, iv
Show Answer & Explanation

Answer: (a) i, ii, iii

Explanation:
1. For identical bodies in a head-on elastic collision, the velocities are exchanged.
2. Hence momenta and kinetic energies are also exchanged.

Teacher's Note:
1. Equal masses swap velocities.
2. Momentum and kinetic energy follow the velocity.

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