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A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in m/s) after collision will be
A
0, 2
B
0, 1
C
1, 1
D
1, 0.5
Detailed Solution
Let the masses be m and 2m and the velocities after collision be $v_1$ and $v_2$.
Conservation of momentum: $m\times2 + 2m\times0 = mv_1 + 2mv_2$, so $v_1 + 2v_2 = 2$ ... (i)
Coefficient of restitution: $e = \frac{v_2 - v_1}{u_1 - u_2}$, so $v_2 - v_1 = 0.5\times(2 - 0) = 1$ ... (ii)
Adding (i) and (ii): $3v_2 = 3$, so $v_2 = 1$ m/s
From (ii): $v_1 = v_2 - 1 = 0$
The velocities after collision are 0 and 1 m/s.
Conservation of momentum: $m\times2 + 2m\times0 = mv_1 + 2mv_2$, so $v_1 + 2v_2 = 2$ ... (i)
Coefficient of restitution: $e = \frac{v_2 - v_1}{u_1 - u_2}$, so $v_2 - v_1 = 0.5\times(2 - 0) = 1$ ... (ii)
Adding (i) and (ii): $3v_2 = 3$, so $v_2 = 1$ m/s
From (ii): $v_1 = v_2 - 1 = 0$
The velocities after collision are 0 and 1 m/s.