JEE Physics System Of Particles and Rotational Motion MCQs Set 07

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Multiple Choice Questions

Question 1: For an inelastic collision between two spherical rigid bodies:
(a) The linear momentum is conserved
(b) The total potential energy is conserved
(c) The total kinetic energy is conserved
(d) The linear momentum is not conserved
Show Answer & Explanation

Answer: (a) The linear momentum is conserved

Explanation:
1. Momentum is conserved in every collision when no external impulse acts.
2. Kinetic energy is lost in an inelastic collision.

Teacher's Note:
1. Momentum conservation never fails for an isolated system. Kinetic energy conservation needs the collision to be elastic.

Question 2: A ball moving with a velocity of 2 m/s collides head-on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, their velocities (in m/s) after the collision will be:
(a) 0, 1
(b) 1, 1
(c) 1, 0.5
(d) 0, 2
Show Answer & Explanation

Answer: (a) 0, 1

Explanation:
1. Masses m and 2m. Momentum: \( 2m = mv_1 + 2mv_2 \Rightarrow v_1 + 2v_2 = 2 \).
2. Restitution: \( v_2 - v_1 = 0.5\times 2 = 1 \).
3. Solving, \( v_2 = 1 \) m/s and \( v_1 = 0 \).

Teacher's Note:
1. Write the two equations (momentum and restitution) and solve them simultaneously.

Question 3: Two particles of masses m₁ and m₂ move with initial velocities u₁ and u₂. On collision, one of the particles gets excited to a higher level after absorbing an energy ε. If the final velocities are v₁ and v₂, then:
(a) \( \tfrac{1}{2}m_1u_1^2 + \tfrac{1}{2}m_2u_2^2 - \varepsilon = \tfrac{1}{2}m_1v_1^2 + \tfrac{1}{2}m_2v_2^2 \)
(b) \( \tfrac{1}{2}m_1u_1^2 + \tfrac{1}{2}m_2u_2^2 = \tfrac{1}{2}m_1v_1^2 + \tfrac{1}{2}m_2v_2^2 - \varepsilon \)
(c) \( m_1^2u_1 + m_2^2u_2 - \varepsilon = m_1^2v_1 + m_2^2v_2 \)
(d) \( \tfrac{1}{2}m_1^2u_1^2 + \tfrac{1}{2}m_2^2u_2^2 + \varepsilon = \tfrac{1}{2}m_1^2v_1^2 + \tfrac{1}{2}m_2^2v_2^2 \)
Show Answer & Explanation

Answer: (a) \( \tfrac{1}{2}m_1u_1^2 + \tfrac{1}{2}m_2u_2^2 - \varepsilon = \tfrac{1}{2}m_1v_1^2 + \tfrac{1}{2}m_2v_2^2 \)

Explanation:
1. The initial kinetic energy equals the final kinetic energy plus the energy absorbed.
2. So initial KE - ε = final KE.

Teacher's Note:
1. Energy absorbed by internal excitation is taken from the kinetic energy. Total energy is conserved.

Question 4: Two spherical bodies of masses M and 5M and radii R and 2R are released in free space with the initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, the distance covered by the smaller body just before collision is:
(a) 2.5R
(b) 1.5R
(c) 4.5R
(d) 7.5R
Show Answer & Explanation

Answer: (d) 7.5R

Explanation:
1. They touch when the centres are 3R apart, so the total approach distance is \( 12R - 3R = 9R \).
2. The centre of mass is fixed, so distances are in the inverse ratio of masses: the smaller body covers \( \dfrac{5}{6}\times 9R = 7.5R \).

Teacher's Note:
1. Keep the centre of mass fixed and divide the total approach distance in the inverse ratio of the masses.

Question 5: A ball is thrown vertically downwards from a height of 20 m with an initial velocity v₀. It collides with the ground, loses 50% of its energy in the collision and rebounds to the same height. The initial velocity v₀ is (g = 10 m/s²):
(a) 20 m/s
(b) 10 m/s
(c) 14 m/s
(d) 28 m/s
Show Answer & Explanation

Answer: (a) 20 m/s

Explanation:
1. Energy just before the bounce: \( \tfrac{1}{2}mv_0^2 + mg(20) \). After the bounce it is half of that and equals \( mg(20) \).
2. \( \tfrac{1}{2}\left(\tfrac{1}{2}v_0^2 + 200\right) = 200 \Rightarrow v_0^2 = 400 \).
3. \( v_0 = 20 \) m/s.

Teacher's Note:
1. The rebound energy fixes the speed after the collision. Work backwards to the energy before it.

Question 6: On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of the same mass M which is initially at rest. After the collision the first block moves at an angle \( \theta \) to its initial direction and has a speed \( v/3 \). The second block's speed after the collision is:
(a) \( \dfrac{3}{4}v \)
(b) \( \dfrac{3}{\sqrt2}v \)
(c) \( \dfrac{\sqrt3}{2}v \)
(d) \( \dfrac{2\sqrt2}{3}v \)
Show Answer & Explanation

Answer: (d) \( \dfrac{2\sqrt2}{3}v \)

Explanation:
1. For identical masses in an elastic oblique collision, the final velocities are perpendicular, so \( v^2 = v_1^2 + v_2^2 \).
2. \( v_2^2 = v^2 - \dfrac{v^2}{9} = \dfrac{8v^2}{9} \Rightarrow v_2 = \dfrac{2\sqrt2}{3}v \).

Teacher's Note:
1. Use the 90° result and energy conservation to get the second speed without the angle.

Question 7: A bullet of mass 10 g moving horizontally with a velocity of 400 m/s strikes a wooden block of mass 2 kg suspended by a light inextensible string of length 5 m. As a result the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges horizontally from the block will be (g = 9.8 m/s²):
(a) 160 m/s
(b) 80 m/s
(c) 100 m/s
(d) 120 m/s
Show Answer & Explanation

Answer: (d) 120 m/s

Explanation:
1. Block speed after the bullet leaves: \( V = \sqrt{2gh} = \sqrt{2\times 9.8\times 0.1} = 1.4 \) m/s.
2. Momentum: \( 0.01\times 400 = 0.01v + 2\times 1.4 \Rightarrow 0.01v = 1.2 \).
3. \( v = 120 \) m/s.

Teacher's Note:
1. Use energy conservation for the block's swing to get its speed, then momentum conservation for the bullet-block collision.

Question 8: Two identical balls A and B having velocities of 0.5 m/s and -0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be:
(a) 0.5 m/s and -0.3 m/s
(b) -0.3 m/s and 0.5 m/s
(c) -0.5 m/s and 0.3 m/s
(d) 0.3 m/s and 0.5 m/s
Show Answer & Explanation

Answer: (a) 0.5 m/s and -0.3 m/s

Explanation:
1. Identical balls exchange velocities in a head-on elastic collision.
2. B takes A's velocity, 0.5 m/s, and A takes B's, -0.3 m/s.

Teacher's Note:
1. The answer is asked in the order B then A. Do not swap them.

Question 9: A wagon of mass 20 metric ton moves with a velocity of 10 m/s and collides perfectly inelastically with a stationary wagon of mass 60 metric ton. The loss of kinetic energy is:
(a) 750 kJ
(b) 500 kJ
(c) 250 kJ
(d) 650 kJ
Show Answer & Explanation

Answer: (a) 750 kJ

Explanation:
1. \( V = \dfrac{20\times 10}{80} = 2.5 \) m/s.
2. \( K_i = \tfrac{1}{2}(20000)(100) = 10^6 \) J; \( K_f = \tfrac{1}{2}(80000)(6.25) = 2.5\times 10^5 \) J.
3. Loss = 750 kJ.

Teacher's Note:
1. Convert tonnes to kilograms (1 metric ton = 1000 kg). The loss is the difference of the two kinetic energies.

Question 10: A moving block of mass m collides with another stationary block of mass 4m. The lighter block comes to rest after the collision. If the initial velocity of the lighter block is v, the coefficient of restitution e is:
(a) 0.25
(b) 0.4
(c) 0.8
(d) 0.5
Show Answer & Explanation

Answer: (a) 0.25

Explanation:
1. Momentum: \( mv = 4mV \Rightarrow V = v/4 \).
2. \( e = \dfrac{V - 0}{v - 0} = \dfrac{v/4}{v} = 0.25 \).

Teacher's Note:
1. Speed of separation divided by speed of approach is e.

Question 11: A body of mass 10 metric ton moving with velocity 5 m/s collides head-on with another body of mass 40 metric ton at rest. They combine and move together. The loss in kinetic energy is:
(a) \( 4\times 10^5 \) J
(b) \( 1\times 10^5 \) J
(c) \( 2\times 10^5 \) J
(d) \( 3\times 10^5 \) J
Show Answer & Explanation

Answer: (b) \( 1\times 10^5 \) J

Explanation:
1. \( V = \dfrac{10\times 5}{50} = 1 \) m/s.
2. \( K_i = \tfrac{1}{2}(10^4)(25) = 1.25\times 10^5 \) J; \( K_f = \tfrac{1}{2}(5\times 10^4)(1) = 2.5\times 10^4 \) J.
3. Loss \( = 1\times 10^5 \) J.

Teacher's Note:
1. Convert tonnes to kg. The fractional loss is 4/5 of the initial energy here.

Question 12: After falling from a height of 5 m, a ball strikes the roof of a lift. At the time of collision the lift is moving upward with a velocity of 1 m/s. The velocity with which the ball rebounds after the collision is (e = 1, g = 10 m/s²):
(a) 12 m/s downwards
(b) 12 m/s upwards
(c) 13 m/s upwards
(d) 11 m/s downwards
Show Answer & Explanation

Answer: (b) 12 m/s upwards

Explanation:
1. Ball speed before: \( \sqrt{2\times 10\times 5} = 10 \) m/s downward; lift moves up at 1 m/s, so the relative speed of approach is 11 m/s.
2. With e = 1 the relative speed of separation is 11 m/s upwards relative to the lift: \( 1 + 11 = 12 \) m/s upwards.

Teacher's Note:
1. Work in the frame of the lift. A perfectly elastic bounce reverses the relative velocity.

Question 13: Two masses m₁ = 2 kg and m₂ = 5 kg are moving on a frictionless surface with velocities 10 m/s and 3 m/s respectively, with m₂ ahead of m₁. An ideal spring of force constant k = 1120 N/m is attached to the back of m₂. The maximum compression of the spring will be:
(a) 0.72 m
(b) 0.25 m
(c) 0.51 m
(d) 0.062 m
Show Answer & Explanation

Answer: (b) 0.25 m

Explanation:
1. At maximum compression both have the same velocity: \( V = \dfrac{2\times 10 + 5\times 3}{7} = 5 \) m/s.
2. Loss of KE \( = \tfrac{1}{2}(2)(100) + \tfrac{1}{2}(5)(9) - \tfrac{1}{2}(7)(25) = 122.5 - 87.5 = 35 \) J.
3. \( \tfrac{1}{2}(1120)x^2 = 35 \Rightarrow x = 0.25 \) m.

Teacher's Note:
1. At maximum compression the relative velocity is zero. The lost kinetic energy is stored in the spring.

Question 14: A frictionless steel ball of radius 2 cm moving on a horizontal plane with a velocity of 5 cm/s collides head-on with another stationary steel ball of radius 3 cm (e = 1). The velocities of the two bodies after the collision are respectively (in cm/s):
(a) -2.7, 2.3
(b) -2.7, -2.3
(c) 2.7, 2.3
(d) 2.7, -2.3
Show Answer & Explanation

Answer: (a) -2.7, 2.3

Explanation:
1. Masses are proportional to \( r^3 \): \( m_1 : m_2 = 8 : 27 \).
2. \( v_1 = \dfrac{8 - 27}{35}\times 5 = -2.71 \); \( v_2 = \dfrac{2\times 8}{35}\times 5 = 2.29 \).

Teacher's Note:
1. Mass is proportional to the cube of the radius for steel balls. Then use the elastic-collision formulae.

Question 15: A neutron makes a head-on elastic collision with a stationary deuteron (mass ratio 1 : 2). The fractional energy loss of the neutron in the collision is:
(a) \( \dfrac{8}{9} \)
(b) \( \dfrac{2}{3} \)
(c) \( \dfrac{8}{27} \)
(d) \( \dfrac{16}{82} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{8}{9} \)

Explanation:
1. Fractional energy transferred \( = \dfrac{4m_1m_2}{(m_1+m_2)^2} = \dfrac{4\times 1\times 2}{9} = \dfrac{8}{9} \).

Teacher's Note:
1. This is why light nuclei are good moderators in a reactor. The formula applies to a stationary target.

Question 16: A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction with one-fourth of its original speed v. The mass of the other body is:
(a) 1.0 kg
(b) 1.2 kg
(c) 1.4 kg
(d) 1.5 kg
Show Answer & Explanation

Answer: (b) 1.2 kg

Explanation:
1. \( v_1 = \dfrac{m_1 - m_2}{m_1 + m_2}v = \dfrac{v}{4} \Rightarrow 4(2 - m_2) = 2 + m_2 \).
2. \( 8 - 4m_2 = 2 + m_2 \Rightarrow m_2 = 1.2 \) kg.

Teacher's Note:
1. Use the elastic formula for a stationary target and solve the linear equation for the unknown mass.

Question 17: A nucleus of mass number A, originally at rest, emits an alpha particle with speed v. The recoil speed of the daughter nucleus is:
(a) \( \dfrac{4v}{A-4} \)
(b) \( \dfrac{v}{A+4} \)
(c) \( \dfrac{4v}{A+4} \)
(d) \( \dfrac{v}{A-4} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{4v}{A-4} \)

Explanation:
1. The daughter has mass number A - 4. Momentum conservation: \( 4v = (A - 4)V \).
2. \( V = \dfrac{4v}{A - 4} \).

Teacher's Note:
1. Use mass numbers as masses. The alpha particle has mass number 4.

Question 18: A particle of mass 2 kg moving at 5 m/s collides head-on with a particle of mass 3 kg moving in the same direction at 4 m/s. The coefficient of restitution is 2/3. The velocities of the two particles after the collision are:
(a) 3 m/s and 5 m/s
(b) 4 m/s and 4.67 m/s
(c) 5 m/s and 4 m/s
(d) 4.5 m/s and 4.5 m/s
Show Answer & Explanation

Answer: (b) 4 m/s and 4.67 m/s

Explanation:
1. Momentum: \( 2(5) + 3(4) = 2v_1 + 3v_2 \Rightarrow 2v_1 + 3v_2 = 22 \).
2. Restitution: \( v_2 - v_1 = \tfrac{2}{3}(5 - 4) = \tfrac{2}{3} \).
3. Solving, \( v_1 = 4 \) m/s and \( v_2 = 4.67 \) m/s.

Teacher's Note:
1. Take the faster body as body 1. The relative speed of approach is 1 m/s, so the relative speed of separation is 2/3 m/s.

Question 19: A block of mass 5 kg moving at 2 m/s collides head-on with a block of mass 3 kg moving at 4 m/s in the opposite direction. The coefficient of restitution is 0.6. The velocities of the 5 kg and the 3 kg blocks after the collision are:
(a) 0 and 1 m/s
(b) -1.6 m/s and 2.0 m/s
(c) 1.6 m/s and -2.0 m/s
(d) -2.5 m/s and 3.5 m/s
Show Answer & Explanation

Answer: (b) -1.6 m/s and 2.0 m/s

Explanation:
1. Take the 5 kg block's direction as positive: \( 5(2) + 3(-4) = 5v_A + 3v_B \Rightarrow 5v_A + 3v_B = -2 \).
2. Restitution: \( v_B - v_A = 0.6\times 6 = 3.6 \).
3. Solving, \( v_A = -1.6 \) m/s and \( v_B = 2.0 \) m/s.

Teacher's Note:
1. Both blocks reverse their directions. With e = 1 the answer would be -2.5 and 3.5 m/s, so the losses are visible by comparing.

Question 20: Two balls of equal masses undergo a head-on collision while moving in opposite directions with speeds of 6 m/s each. If the coefficient of restitution is 1/3, the speed of each ball after the impact is:
(a) 6 m/s
(b) 3 m/s
(c) 4 m/s
(d) 2 m/s
Show Answer & Explanation

Answer: (d) 2 m/s

Explanation:
1. The total momentum is zero, so after the collision the speeds are equal and opposite.
2. Relative speed of approach = 12 m/s, so relative speed of separation \( = 12/3 = 4 \) m/s, i.e. 2 m/s each.

Teacher's Note:
1. In the centre-of-mass frame both balls simply reverse and the speeds are multiplied by e.

Question 21: A body of mass 1 kg moving with a velocity of 1 m/s makes a head-on collision with an identical stationary body. The force of interaction increases linearly from zero to F₀ in 0.5 s and then decreases linearly to zero in a further 0.5 s. If the first body is brought to rest, the magnitude of F₀ is:
(a) 1 N
(b) 4 N
(c) 2 N
(d) 0.5 N
Show Answer & Explanation

Answer: (c) 2 N

Explanation:
1. The impulse is the area under the force-time triangle: \( \tfrac{1}{2}F_0\times 1 \).
2. The impulse equals the change in momentum \( = 1\times 1 = 1 \) N s, so \( F_0 = 2 \) N.

Teacher's Note:
1. The area of a triangle is half base times height. Equate the impulse to the momentum change.

Question 22: A 1 kg ball is dropped from a height of 20 m and rebounds after each collision with the ground with e = 0.5. The velocity of the ball just after the second collision is (g = 10 m/s²):
(a) 10 m/s
(b) 5 m/s
(c) 2.5 m/s
(d) 20 m/s
Show Answer & Explanation

Answer: (b) 5 m/s

Explanation:
1. Speed just before the first collision \( = \sqrt{2gh} = 20 \) m/s.
2. After the first collision 10 m/s, after the second \( 20\times e^2 = 5 \) m/s.

Teacher's Note:
1. Each collision multiplies the speed by e. After n collisions the speed is e to the power n times the original.

Question 23: A ball is thrown vertically upward from the ground with a speed of 40 m/s. It collides with the ground after returning, with e = 0.5. The total distance travelled by the ball before it comes to rest is (g = 10 m/s²):
(a) 160 m
(b) 133.3 m
(c) 213.3 m
(d) 80 m
Show Answer & Explanation

Answer: (c) 213.3 m

Explanation:
1. The maximum height of the first flight \( h = 40^2/20 = 80 \) m; going up and coming down covers 160 m.
2. After this the ball falls from 80 m repeatedly: the later distance \( = 2h\dfrac{e^2}{1-e^2} = 160\times\dfrac{0.25}{0.75} = 53.3 \) m.
3. Total \( = 160 + 53.3 = 213.3 \) m.

Teacher's Note:
1. The first 'up and down' is a full 2h. Subsequent trips form a geometric series.

Question 24: Two particles of masses m and m moving perpendicular to each other with the same speed 45√2 m/s stick together on collision. The velocity of the combined mass after collision is:
(a) \( 45\sqrt2 \) m/s
(b) 45 m/s
(c) 22.5 m/s
(d) 90 m/s
Show Answer & Explanation

Answer: (b) 45 m/s

Explanation:
1. The momenta \( mv \) are perpendicular: the resultant is \( \sqrt2\,mv = \sqrt2\,m\times 45\sqrt2 = 90m \).
2. The combined mass is 2m, so \( V = 90m/2m = 45 \) m/s.

Teacher's Note:
1. Vector addition of the momenta is the key step. Divide the resultant by the total mass.

Question 25: A particle of mass 2 kg moving with velocity 5 m/s along +x collides head-on with a 3 kg particle moving at 2 m/s along -x. After the collision the first particle has a speed of 1.6 m/s in the negative x direction. The coefficient of restitution is:
(a) \( \dfrac{5}{7} \)
(b) \( \dfrac{4}{7} \)
(c) \( \dfrac{3}{7} \)
(d) \( \dfrac{2}{7} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{4}{7} \)

Explanation:
1. \( 2(5) + 3(-2) = 2(-1.6) + 3v_2 \Rightarrow 4 = -3.2 + 3v_2 \Rightarrow v_2 = 2.4 \) m/s.
2. \( e = \dfrac{2.4 - (-1.6)}{5 - (-2)} = \dfrac{4}{7} \).

Teacher's Note:
1. Find the second velocity from momentum and then use the definition of e.

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