JEE Physics System Of Particles and Rotational Motion MCQs Set 57

Answer: (c) 0.89 and 0.28

Explanation:
1. Fractional loss in an elastic head-on collision \( =\dfrac{4m_1m_2}{(m_1+m_2)^2} \).
2. Deuteron: \( \dfrac{4\cdot2}{9}=0.89 \). Carbon: \( \dfrac{4\cdot12}{169}=0.28 \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (d) 56 %

Explanation:
1. Initial KE \( =\tfrac12m(2v)^2+\tfrac12(2m)v^2=3mv^2 \).
2. Momentum: \( (2mv,\,2mv) \), so \( V=\dfrac{2\sqrt2\,mv}{3m} \) and final KE \( =\tfrac12(3m)\dfrac{8v^2}9=\dfrac43mv^2 \).
3. Loss \( =\dfrac{3-4/3}{3}=\dfrac59\approx56\% \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (d) A is false, but R is true.

Explanation:
1. Statement I is false: for a perfectly inelastic collision \( V=\dfrac{mv}{M+m} \), and the loss is \( \tfrac12mv^2\left(1-\dfrac m{M+m}\right)=\tfrac12mv^2\cdot\dfrac M{M+m} \), so \( f=\dfrac{M}{M+m} \).
2. Statement II is true: the loss is greatest when the bodies move together after the collision.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (b) 500 m/s

Explanation:
1. Time of fall \( =\sqrt{2(5)/10}=1 \) s, so the ball's speed is 20 m/s and the bullet's is 100 m/s.
2. \( 0.01v=0.2(20)+0.01(100)=5\Rightarrow v=500 \) m/s.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (a) \( 2(m_1+m_2)\,g\,t_0 \)

Explanation:
1. The collision forces are internal, so the total momentum changes only because of gravity, at the rate \( (m_1+m_2)g \) over the whole time from 0 to \( 2t_0 \).
2. This gives \( |\Delta\vec P|=(m_1+m_2)g\cdot2t_0 \). Since t0 is the time of the collision and the interval from 0 to 2t0 is 2t0 long, the answer is \( 2(m_1+m_2)gt_0 \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (d) y = −5 cm

Explanation:
1. The centre of mass keeps moving along the x-axis (y = 0) since no force acts in the horizontal plane.
2. \( \dfrac m4(15)+\dfrac{3m}4y=0\Rightarrow y=-5 \) cm.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (a) \( 3v\cos\theta \)

Explanation:
1. At the top the speed is \( v\cos\theta \) (horizontal). One piece retraces its path: velocity \( -v\cos\theta \).
2. \( 2m\,v\cos\theta=m(-v\cos\theta)+mu\Rightarrow u=3v\cos\theta \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (b) the total momentum of the ball and the earth is conserved

Explanation:
1. The ball and the earth form an isolated system, so their total momentum is conserved.
2. The ball's own momentum changes (the earth supplies an impulse), and an inelastic collision loses mechanical energy of the ball-earth system.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (c) 30

Explanation:
1. \( H=\dfrac{u^2\sin^245^\circ}{2g}=\dfrac{u^2}{4g}=120\Rightarrow u^2=480g \).
2. After the bounce \( v^2=u^2/2=240g \) and the angle is 30°.
3. \( H'=\dfrac{v^2\sin^230^\circ}{2g}=\dfrac{240g\times\tfrac14}{2g}=30 \) m.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Statement I is true: for an elastic collision the coefficient of restitution is 1, so the speed of separation equals the speed of approach.
2. Statement II is also true, but momentum conservation holds for every collision; it does not by itself explain the equality of relative speeds (that needs the energy conservation too).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (a) \( \sqrt{45}\ \text{m/s} \)

Explanation:
1. Speed just before reaching B: \( v^2=2g(3)=60 \).
2. The velocity makes \( 60^\circ-30^\circ=30^\circ \) with the second incline. In a totally inelastic collision the component normal to the incline is lost, leaving \( v\cos30^\circ \).
3. \( v_B'^2=60\cdot\tfrac34=45 \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (b) \( \sqrt{105}\ \text{m/s} \)

Explanation:
1. After B: \( v_B'^2=45 \). The block slides down a further 3 m vertically on a smooth incline.
2. \( v_C^2=45+2g(3)=105 \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (c) zero

Explanation:
1. The incoming velocity makes 30° with the second incline. An elastic collision reverses the normal component and keeps the tangential component, so the velocity turns by 60° from the first incline direction.
2. The new direction is \( -60^\circ+60^\circ=0^\circ \): horizontal. Hence the vertical component is zero.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (d) (a) and (c) only

Explanation:
1. \( \dfrac{1-5}{6}u=-2\Rightarrow u=3 \) m/s: total momentum \( =3 \) kg m/s (a true). The 5 kg mass gets \( \dfrac{2(1)}{6}(3)=1 \) m/s, momentum 5 kg m/s, so (b) is false.
2. \( v_{cm}=3/6=0.5 \) m/s and \( KE_{cm}=\tfrac12(6)(0.25)=0.75 \) J: (c) true. Total KE \( =\tfrac12(1)(9)=4.5 \) J, so (d) is false.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (a) \( \dfrac32mv^2 \)

Explanation:
1. The third fragment has mass 2m. Momentum: \( 2m\,u=\sqrt2\,mv\Rightarrow u=\dfrac{v}{\sqrt2} \).
2. Energy \( =\tfrac12mv^2+\tfrac12mv^2+\tfrac12(2m)\dfrac{v^2}2=\dfrac32mv^2 \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.