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

Model answer

Also accepted: 2 m/s, 2.0, 2 ms^-1

Explanation

Take A's original direction as positive and let the speeds after the impact be

v_A

and

v_B

. Conservation of momentum gives

2(4) = 2v_A + 3v_B

, so

2v_A + 3v_B = 8

. Newton's law of restitution gives the speed of separation as

\dfrac14

of the speed of approach:

v_B - v_A = \dfrac14(4) = 1

. Substituting

v_A = v_B - 1

gives

5v_B = 10

, so

v_B = 2\,\mathrm{ms^{-1}}

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

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