Impulse & Momentum

We have s = $ut + \frac{1}{2} a t^2$ In vertical direction, s = 5 m, u = 0 and a = g = 10 m/s. Therefore, 5m = $\frac{1}{2} 10 t^2$ ⇒ t = 1s For the ball, the horizontal displacement is 20 m. Therefore, 20m = $v_{\text{ball}}$ × t ⇒ $v_{\text{ball}}$ = 20 m/s or $P_{\text{ball}}$ = 0.2 kg × 20 m/s = 4 kg m/s Similarly, $P_{\text{bullet}}$ = 1 kg m/s Therefore, by conservation of momentum, we have that the momentum of bullet before collision as follows: $P_{\text{ball}}$ + $P_{\text{bullet}}$ ⇒ 0.01 kg × u = 5 kg m/s or ⇒ u = $\frac{5}{0.01}$ = 500 m/s