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NVAMediumJEE 2023Elastic & Inelastic Collisions

JEE Physics 2023 Question with Solution

A body of mass 1 kg1 \, \text{kg} collides head-on with a stationary body of mass 3 kg3 \, \text{kg}. After the collision, the smaller body reverses its direction of motion and moves with a speed of 2 m/s2 \, \text{m/s}. The initial speed of the smaller body before collision is:

Answer

Correct answer:4

Step-by-step solution

Standard Method

Given: mass of smaller body m1=1 kgm_1 = 1 \, \text{kg}, mass of stationary body m2=3 kgm_2 = 3 \, \text{kg}, initial velocity of second body u2=0u_2 = 0, and after collision the smaller body reverses direction with speed 2 m/s2 \, \text{m/s}.

Find: the initial speed u1u_1 of the smaller body.

Using conservation of momentum,

m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

Substituting the given values and taking the reversed direction of the smaller body as negative,

1×u1=−2+3v1 \times u_1 = -2 + 3v

So,

u1=−2+3vu_1 = -2 + 3v

From the relation shown in the solution figure,

1=v+2u11 = \frac{v+2}{u_1}

Hence,

u1=v+2u_1 = v + 2

Solving these equations,

u1=4 m/su_1 = 4 \, \text{m/s}

Therefore, the initial speed of the smaller body is 4 m/s4 \, \text{m/s}.

Collision diagram showing a 1 kg body moving toward a stationary 3 kg body, then after collision the 1 kg body moving backward at 2 m/s and the 3 kg body moving forward with speed v.

Using momentum conservation

The solution states that by applying the principle of conservation of momentum and considering the reversal of direction after collision, the initial speed is obtained as 4 m/s4 \, \text{m/s}. The correct answer is 44.

Common mistakes

  • Taking the final velocity of the smaller body as positive 2 m/s2 \, \text{m/s}. This is wrong because the body reverses its direction after collision. Use −2 m/s-2 \, \text{m/s} with a consistent sign convention.

  • Ignoring that the second body was initially stationary. This leads to an unnecessary momentum term. Since it is stationary initially, use u2=0u_2 = 0.

  • Using only speeds instead of velocities in the momentum equation. Momentum conservation is vector-based in one dimension, so direction must be included through signs.

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