JEE Physics System Of Particles and Rotational Motion MCQs Set 18

Answer: (a) \( 40\sqrt{2} \) m/s

Explanation:
1. Masses are \( \dfrac{4}{15}, \dfrac{4}{15}, \dfrac{7}{15} \) kg.
2. The resultant momentum of the two equal parts is \( \dfrac{4}{15}(70)\sqrt{2} \).
3. The bigger part has \( \dfrac{7}{15}v = \dfrac{4}{15}(70)\sqrt{2} \Rightarrow v = 40\sqrt{2} \) m/s.

Teacher's Note:
1. Only the ratio of masses matters, not the total mass.
2. The third momentum must cancel the vector sum of the other two.

Answer: (d) unitless

Explanation:
1. \( e = \dfrac{\text{velocity of separation}}{\text{velocity of approach}} \), a ratio of two speeds.
2. So it has no unit.

Teacher's Note:
1. It is a pure number between 0 and 1.
2. Perfectly elastic gives 1.

Answer: (a) \( \dfrac{M - m}{M}u \)

Explanation:
1. Momentum conservation: \( Mu = Mv + mu \).
2. \( v = \dfrac{(M - m)u}{M} \).

Teacher's Note:
1. Write conservation of momentum and solve for the unknown.
2. Option (d) is circular because it contains v itself.

Answer: (c) 9 kg

Explanation:
1. \( 3v = (3 + m)\dfrac{v}{4} \Rightarrow 3 + m = 12 \).
2. So \( m = 9 \) kg.

Teacher's Note:
1. After the collision the two bodies move together with a common velocity.
2. Cancel v from both sides.

Answer: (a) \( \dfrac{1 - e}{1 + e} \)

Explanation:
1. Momentum: \( v_1 + v_2 = u \). Restitution: \( v_2 - v_1 = eu \).
2. \( v_1 = \dfrac{(1 - e)u}{2} \) and \( v_2 = \dfrac{(1 + e)u}{2} \).
3. \( \dfrac{v_1}{v_2} = \dfrac{1 - e}{1 + e} \).

Teacher's Note:
1. Use the two equations of momentum and restitution together.
2. For e = 1 the first sphere stops, as expected.

Answer: (b) 0.2

Explanation:
1. Take A's direction as positive. Momentum: \( 1(21) + 2(-4) = 1(1) + 2v_B \Rightarrow v_B = 6 \) m/s.
2. \( e = \dfrac{v_B - v_A}{u_A - u_B} = \dfrac{6 - 1}{21 + 4} = 0.2 \).

Teacher's Note:
1. Velocity of separation divided by velocity of approach.
2. Apply signs to the directions carefully.

Answer: (a) 15 and 20

Explanation:
1. Equal masses in an elastic head-on collision exchange velocities.
2. P gets 15 m/s and Q gets 20 m/s.

Teacher's Note:
1. Exchange of velocity is a handy shortcut.
2. It applies to elastic, one-dimensional collisions of equal masses.

Answer: (c) \( m = M \)

Explanation:
1. After the collision \( v_1 = \dfrac{(m - M)v}{m + M} \).
2. This is zero only if \( m = M \).

Teacher's Note:
1. Equal masses exchange velocities in elastic head-on collisions.
2. The moving body stops and the other starts.

Answer: (b) mgL

Explanation:
1. The bob reaches the bottom with kinetic energy mgL.
2. Equal masses in an elastic collision exchange velocities, so the block takes all of it.

Teacher's Note:
1. Elastic equal-mass collision transfers all the kinetic energy.
2. The bob stops dead.

Answer: (b) 7 : 5

Explanation:
1. Let \( m_1 = m, u_1 = 3u \) and \( m_2 = 2m, u_2 = -u \).
2. \( v_1 = \dfrac{(m - 2m)3u + 2(2m)(-u)}{3m} = -\dfrac{7u}{3} \) and \( v_2 = \dfrac{(2m - m)(-u) + 2m(3u)}{3m} = \dfrac{5u}{3} \).
3. The speed ratio is 7 : 5.

Teacher's Note:
1. Use the standard elastic collision formulas with signs.
2. Check that momentum and relative speed are preserved.

Answer: (b) \( \dfrac{1}{3} \)

Explanation:
1. \( v_A = \dfrac{(m_A - m_B)u}{m_A + m_B} \) and \( v_B = \dfrac{2m_A u}{m_A + m_B} \).
2. Equal speeds in opposite directions: \( v_A = -v_B \Rightarrow m_B - m_A = 2m_A \).
3. \( \dfrac{m_A}{m_B} = \dfrac{1}{3} \).

Teacher's Note:
1. Set the two final velocities equal and opposite.
2. The lighter body A bounces back.

Answer: (c) 60 m/s

Explanation:
1. For a very heavy body hitting a light body at rest elastically, the light body's velocity is \( v_2 = \dfrac{2m_1 u}{m_1 + m_2} \approx 2u = 60 \) m/s.
2. The heavy body continues almost unchanged.

Teacher's Note:
1. Think of the light body as bouncing off a moving wall.
2. The result is twice the heavy body's speed.

Answer: (c) The momentum is conserved in an elastic collision only.

Explanation:
1. Momentum is conserved in every collision, elastic or inelastic, when no external force acts.
2. So (c) is wrong.
3. A body at rest can have potential energy and no momentum, so (b) is right, and a body with zero potential energy can still move, so (a) is right.

Teacher's Note:
1. Momentum conservation does not depend on the type of collision.
2. Only kinetic energy depends on it.

Answer: (c) Momentum is conserved in all collisions but not kinetic energy.

Explanation:
1. Momentum is conserved in all collisions (no external force).
2. Kinetic energy is conserved only in elastic collisions, not in all collisions.

Teacher's Note:
1. Statement (a) mixes up elastic and inelastic.
2. Kinetic energy is lost in inelastic collisions.

Answer: (d) Velocity of separation may be greater than velocity of approach.

Explanation:
1. In an inelastic collision the velocity of separation is less than the velocity of approach (e < 1).
2. So (d) is not correct.
3. (a) happens in a perfectly inelastic collision, and (b) and (c) are true.

Teacher's Note:
1. The coefficient of restitution is at most 1 for any collision.
2. Greater than 1 would add energy.

Answer: (b) \( \dfrac{PQ}{P + R} \)

Explanation:
1. Momentum conservation: \( PQ = (P + R)v \).
2. \( v = \dfrac{PQ}{P + R} \).

Teacher's Note:
1. The bullet and body move together after the collision.
2. Check that the answer has the dimension of velocity.

Answer: (c) \( \dfrac{m}{M + m}v \)

Explanation:
1. The impact is very short, so the string tension does not change the horizontal momentum.
2. \( mv = (M + m)V \Rightarrow V = \dfrac{m}{M + m}v \).

Teacher's Note:
1. Apply momentum conservation just before and after the impact.
2. Energy is not conserved in the embedding.

Answer: (b) 0.4 m/s

Explanation:
1. \( 0.02(400) = (20 + 0.02)V \).
2. \( V = \dfrac{8}{20.02} \approx 0.4 \) m/s.

Teacher's Note:
1. Convert 20 g to 0.02 kg.
2. The bullet's mass is negligible next to the ice in the denominator.

Answer: (a) 10 cm/s towards east

Explanation:
1. Taking east as positive: \( 10(20) - 4(15) = 14\,V \).
2. \( V = \dfrac{140}{14} = 10 \) cm/s towards east.

Teacher's Note:
1. Treat opposite directions with opposite signs.
2. The heavier momentum decides the final direction.

Answer: (b) 160 J

Explanation:
1. \( 16(5) = 20\,V \Rightarrow V = 4 \) m/s.
2. \( KE = \tfrac{1}{2}(20)(16) = 160 \) J.

Teacher's Note:
1. Velocity from momentum conservation first.
2. The initial KE was 200 J, so 40 J was lost.

Answer: (d) 95%

Explanation:
1. Common velocity \( V = \dfrac{50(10)}{1000} = 0.5 \) m/s.
2. Final KE \( = \tfrac{1}{2}(1)(0.25) = 0.125 \) J and initial KE \( = \tfrac{1}{2}(0.05)(100) = 2.5 \) J.
3. Fraction left \( = 5\% \), so the loss is 95%.

Teacher's Note:
1. Fraction of energy retained in a perfectly inelastic collision is \( m/(m + M) \).
2. Here it is 0.05.

Answer: (c) \( \dfrac{\ell}{4} \)

Explanation:
1. Speed at the bottom \( v = \sqrt{2g\ell} \).
2. After sticking, \( V = \dfrac{v}{2} \).
3. Height \( h = \dfrac{V^2}{2g} = \dfrac{v^2}{8g} = \dfrac{\ell}{4} \).

Teacher's Note:
1. The common velocity is half the original speed.
2. Height varies as the square of the speed.

Answer: (a) 0.62

Explanation:
1. \( e = \sqrt{\dfrac{h_2}{h_1}} = \sqrt{\dfrac{25}{64}} = \dfrac{5}{8} = 0.625 \approx 0.62 \).

Teacher's Note:
1. The mass is irrelevant.
2. Take the square root of the height ratio.

Answer: (a) 10 m

Explanation:
1. Energy at the ground \( = mg(20) \).
2. Half is lost, so \( mgh' = \tfrac{1}{2}mg(20) \Rightarrow h' = 10 \) m.

Teacher's Note:
1. Height is proportional to the energy kept.
2. Half the energy means half the height.

Answer: (c) \( \dfrac{1}{\sqrt{2}} \)

Explanation:
1. The loss of kinetic energy is \( \tfrac{1}{2}\mu u^2(1 - e^2) \) with \( \mu = \dfrac{m}{2} \), so the fraction lost is \( \dfrac{1 - e^2}{2} \).
2. \( \dfrac{1 - e^2}{2} = \dfrac{1}{4} \Rightarrow e^2 = \dfrac{1}{2} \).
3. \( e = \dfrac{1}{\sqrt{2}} \).

Teacher's Note:
1. The loss depends on the reduced mass and the relative speed.
2. Check with \( e = 1 \), where the loss is zero.

Answer: (c) \( 90^\circ \)

Explanation:
1. Momentum gives \( \vec{v}_i = \vec{v}_1 + \vec{v}_2 \), and energy gives \( v_i^2 = v_1^2 + v_2^2 \).
2. Together these require \( \vec{v}_1\cdot\vec{v}_2 = 0 \), so the angle is 90°.

Teacher's Note:
1. Equal-mass elastic oblique collisions with a stationary target always give a right angle.
2. This is the billiard-ball result.

Answer: (b) \( \dfrac{5}{3} \) m/s (about 1.67 m/s)

Explanation:
1. The momenta are \( 40\times 2 = 80 \) and \( 20\times 3 = 60 \), perpendicular to each other.
2. Resultant momentum \( = \sqrt{80^2 + 60^2} = 100 \) kg m/s.
3. \( V = \dfrac{100}{60} = \dfrac{5}{3} \) m/s.

Teacher's Note:
1. Add perpendicular momenta by Pythagoras.
2. Divide by the total mass of 60 kg.

Answer: (c) \( \dfrac{v}{\sqrt{2}} \)

Explanation:
1. Momentum east \( = mv \) and north \( = mv \), so the resultant is \( \sqrt{2}\,mv \) towards north-east.
2. \( 2m\,V = \sqrt{2}\,mv \Rightarrow V = \dfrac{v}{\sqrt{2}} \).

Teacher's Note:
1. Add the momenta as vectors.
2. Equal components give the 45° direction.