A smooth sphere A of mass $2$ kg moving at $4\,\mathrm{ms^{-1}}$ collides directly with a stationary smooth sphere B of mass $3$ kg, the coefficient of restitution being $\dfrac14$ . After the impact B moves off at $2\,\mathrm{ms^{-1}}$ . Find the speed of A immediately after the impact, in $\mathrm{ms^{-1}}$ .

Model answer

Also accepted: 1 m/s, 1.0, 1 ms^-1

Explanation

Conservation of momentum along the line of centres gives

2(4) = 2v_A + 3(2)

, so

2v_A = 8 - 6 = 2

and

v_A = 1\,\mathrm{ms^{-1}}

. As a check, the law of restitution requires the separation speed

v_B - v_A = 2 - 1 = 1

to equal

\dfrac14

of the approach speed

4

, which it does.

Derived from ZIMSEC Additional Mathematics Paper 2, Specimen Paper, Q5

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