JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Conservation of Momentum

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A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body?

(A)1.5 kg
(B)1.2 kg✓ Correct
(C)1.8 kg
(D)1.0 kg

Answer: (B)

Solution

Let the 2 kg body start with speed

v_{0}

and the target have mass

m

. Momentum conservation gives

2 v_{0} = 2 \cdot \frac{v _{0}}{4} + m v

, so

m v = \frac{3 v _{0}}{2}

. For an elastic collision the speed of separation equals the speed of approach,

v - \frac{v _{0}}{4} = v_{0}

, so

v = \frac{5 v _{0}}{4}

. Substituting,

\frac{3 v _{0}}{2} = m \cdot \frac{5 v _{0}}{4}

, which gives

m = \frac{12}{10} = 1.2

kg.

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option

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