Question

A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 426 m/s. (a) What is the resulting speed of the block? (b) What is the speed of the bullet-block center of mass?

Answer 1

Answer:

a) $V_{wf}$ = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

$m_{b}V{b_{i} } + V_{wi} = m_{w} V_{wf} + m_{b}V_{bf}$

where $m_{b},V{b_{i}$ are the mass and the initial velocity of the bullet, $m_{w}$ and $V_{wi}$ are the mass and the initial velocity of the wooden block, and $V_{wf}$ and $V_{bf}$ are the final velocities of the wooden block and the bullet

The wooden block is initial at rest $(V_{wi} = 0)$ this yields

$m_{b}V{b_{i} } = m_{w} V_{wf} + m_{b}V_{bf}$

By solving for $V_{wf}$ adn substitute the givens

$V_{wf}$ = $\frac{m_{b}V_{bi} - m_{b} V_{bf} }{m_{W} }$

= $\frac{5.3(g)(963(m/s)-426(m/s) }{610(g)}$

$V_{wf}$ = 4.67m/s

b) The center of mass speed is defined as

$V = \frac{m_{b} }{m_{b}+m_{w} } V_{bi}$

substituting:

$V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)$

V = 8.29 m/s