A 45.0g bullet is fired on a 2.0kg wooden block which is resting on a horizontal surface with a coefficient friction of 0.4 . If the speed of the bullet is 165m/s when it hits the block. Determine the maximum horizontal distance of the block will reach
2 answers
oh yea the bullet was embedded to the block after it hits
momentum before = momentum just after
.045 * 165 = 2.045 * Vi
solve for Vi, the inital velocity fater collision
Ke = (1/2)(2.045)(Vi^2) initial kinetice energy of bullet-block system
That Ke will be the work done by friction:
Ke = F * d = mu m g * d
m cancels so
so
0.4 (9.81) d = 0.5 Vi^2
.045 * 165 = 2.045 * Vi
solve for Vi, the inital velocity fater collision
Ke = (1/2)(2.045)(Vi^2) initial kinetice energy of bullet-block system
That Ke will be the work done by friction:
Ke = F * d = mu m g * d
m cancels so
so
0.4 (9.81) d = 0.5 Vi^2