Asked by gie
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
Answers
Answered by
gie
oh yea the bullet was embedded to the block after it hits
Answered by
Damon
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
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