Asked by James
A 5.88 g bullet is fired vertically into a 0.9 kg block of wood. The bullet gets stuck in the block, and the impact lifts the block 0.87 m up. (That is, the
block — with the bullet stuck in it — rises 0.87 m up above its initial position, and then falls back down.)
Given g = 9.8 m/s2. What was the initial velocity of the bullet?
Answer in units of m/s.
block — with the bullet stuck in it — rises 0.87 m up above its initial position, and then falls back down.)
Given g = 9.8 m/s2. What was the initial velocity of the bullet?
Answer in units of m/s.
Answers
Answered by
bobpursley
the KE of the block/bullet must be equal to the change in PE of the block /bullet
1/2 (M+m)v1^2=(m+M)gh
solve for V1^2, the initial velocity of the block/bullet combination.
but at impact of the bullet, conservation of momentum applies
mVbullet= (m+M)V1
solve for vbullet.
1/2 (M+m)v1^2=(m+M)gh
solve for V1^2, the initial velocity of the block/bullet combination.
but at impact of the bullet, conservation of momentum applies
mVbullet= (m+M)V1
solve for vbullet.
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