Question:

The track shown in the figure is frictionless. The block B of mass 4 kg is lying at rest and block A of mass 2 kg is pushed along the track with some speed. The collision between A and B is perfectly elastic.With what velocity should the block A start such that block B just reaches at point P?

JEE Main

Updated On: Mar 31, 2026

(A) 10 m/s
(B) 5 m/s
(C) 20 m/s
(D) 15 m/s

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The Correct Option is A

Approach Solution - 1

Explanation:
Given:Mass of the block A, mA = 2 kgMass of the block B, mB = 4 kgHeight of block A from the ground, = 4 mThe collision between block A and block B is perfectly elastic.We have to find the velocity with which the block A should start such that block B just reaches the point P.Let the block A starts with velocity uA and it reaches to block B with the velocity vA.Applying the law of conservation of energy for block A, we have

\frac{1}{2} m_{A} u_{A}^{2} + m_{A} g h = \frac{1}{2} m_{A} v_{A}^{2}

\Rightarrow \frac{μ^{2} A}{2} + 10 \times 4 = \frac{v^{2} A}{2}

\Rightarrow \frac{μ_{A}^{2} + 80}{2} = \frac{v_{A}^{2}}{2}

\Rightarrow v_{A}^{2} = μ_{A}^{3} + 80

....(i)This is the velocity with which block A collides with the block B.Applying the law of conservation of linear momentum just before and after the collision, we have

m_{A} v_{A} + 0 = m_{A} v_{A} + m_{B} u_{B},

[Initial momentum of the block B is zero as it is at rest.]where,VA : final velocity of the block AUB : velocity of the block B after the collision

\Rightarrow 2 v_{A} = 2 V_{A} = 4 u_{B}

\Rightarrow v_{A} = 2 u_{B} + V_{A}

.......(ii)Now, as the collision between the blocks A and B is elastic, applying the law of conservation of kinectic energy just before and after the collision. We have,

\frac{1}{2} m_{A} v_{A}^{2} = \frac{1}{2} m_{A} v_{A}^{2} + \frac{1}{2} m_{B} u_{B}^{2}

\Rightarrow v_{A}^{2} = 2 u_{B}^{2} + V_{A}^{2}

.....(iii)Substituting

v_{A}

from eq. (iii) in eq. (ii), we get

v_{A} = 2 u_{B} + \sqrt{v_{A}^{2} - 2 u_{B}^{2}}

\Rightarrow v_{A} - 2 u_{B} = \sqrt{v_{A}^{2} - 2 u_{B}^{2}}

\Rightarrow v_{A}^{2} + 4 u_{B}^{2} - 4 v_{A} u_{B} = v_{A}^{2} - 2 u_{B}^{2}

[squaring both sides]

\Rightarrow 4 u_{B}^{2} + 2 u_{B}^{2} = 4 v_{A} u_{B}

\Rightarrow 6 u_{B}^{2} = 4 v_{A} u_{B}

\Rightarrow 6 u_{B} = 4 v_{A}

\Rightarrow u_{B} = \frac{2}{3} v_{A}

.....(iv)Substituting

v_{A}

from eq. (i) in eq. (iv), we get

u_{B} = \frac{2}{3} \sqrt{u_{A}^{2} + 80}

......(v)This is the velocity of the block B with which it moves after the collision.To reach just at the point P, all of the kinetic energy of the block B gets converted into its poteintial energy .

\Rightarrow \frac{1}{2} m_{B} u_{B}^{2} = m_{B} g h

\Rightarrow u_{B}^{2} = 2 g h

\Rightarrow \frac{4}{9} (u_{A}^{2} + 80) = 2 \times 10 \times 4

.......[using eq .(v)]

\Rightarrow u_{A}^{2} + 80 = \frac{2 \times 10 \times 4 \times 9}{4}

\Rightarrow u_{A}^{2} = 100 m^{2} / s^{2}

\Rightarrow u_{A} = 10 m / s

This is the velocity with which the block A should start.Hence, the correct option is (A).

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Approach Solution -2

The interaction between the two bodies due to which the direction and magnitude of the velocity of the colliding bodies changes are called a collision.

A collision occurs when two bodies come in direct contact with each other.
It is a situation in which two or more bodies exert forces on each other in about a relatively short time.

Elastic Collision

If in a particular collision, there is no dissipation of energy, the total kinetic energy of the objects before collision is equal to the total kinetic energy of the objects after collision. Such a collision is termed an Elastic collision.

A collision between two particles is said to be a perfectly elastic collision if both the linear momentum and kinetic energy of the system remain conserved.
The value of the coefficient of restitution in a perfectly elastic collision is, e = 1

Inelastic Collision

If, in a particular collision, there is a dissipation of energy, the total kinetic energy of the objects before and after collision is not conserved. Such a collision is termed an inelastic collision.

A collision is said to be a perfectly inelastic collision if two colliding bodies after collision stick together and moves as one body.
In this collision, the linear momentum of the system remains conserved but the kinetic energy is not conserved.
The value of the coefficient of restitution in a perfectly inelastic collision is, e = 0.

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