Question

A 0.05 kg arrow with a velocity of 180m/s is used to shot a 5 kg block of wood off the top of a log. Whats the velocity of the arrow/log after contact?

Answer 1

Question #14630

Let $m_1, v_0$ note the mass and initial velocity of arrow respectively. $m_2$ Is the mass of the block. Use the conservation of momentum and energy ( $v_1, v_2$ are the velocities after collision of arrow and block respectively):

Momentum: $m_{1}v_{0} = m_{1}v_{1} + m_{2}v_{2}$ (1)

Energy: $m_{1}v_{0}^{2} = m_{1}v_{1}^{2} + m_{2}v_{2}^{2}$ (2)

From (1), $v_{1} = v_{0} - \frac{m_{2}}{m_{1}} v_{2}$ , and substituting it into (2), obtain $v_{2}[-2\frac{m_{2}}{m_{1}} v_{0} + \left(\frac{m_{2}}{m_{1}}\right)^{2} v_{2} + m_{2} v_{2}] = 0$

, which gives $v_{2} = \frac{2\frac{m_{2}}{m_{1}}v_{0}}{\left[m_{2} + \left(\frac{m_{2}}{m_{1}}\right)^{2}\right]}\approx 3.6~m / s$

For arrow, $v_{1} = v_{0} - \frac{m_{2}}{m_{1}} v_{2} = -180 \, \text{m/s}$ (At the moment of collision it changes the speed to the opposite, and gives some of its impulse to block).