A 1.95-kg wooden block rests on a table over a large hole as in the figure below

Question

A 1.95-kg wooden block rests on a table over a large hole as in the figure below. A 4.50-g bullet with an initial velocity vi is fired upward into the bottom of the block and remains in the block after the collision. The block and bullet rise to a maximum height of18.0 cm.

Calculate the initial velocity of the bullet from the information provided. (Let up be the positive direction.)

A 1.95-kg wooden block rests on a table over a large hole as in the figure below. A 4.50-g bullet with an initial velocity vi is fired upward into the bottom of the block and remains in the block after the collision. The block and bullet rise to a maximum height of18.0 cm. Calculate the initial velocity of the bullet from the information provided. (Let up be the positive direction.)

Explanation / Answer

m = 4.5 g = 0.0045 kg

M = 1.95 kg

h = 18 cm = 0.18 m

initial momentum of the bullet before collision, pi = m*vi

After collision, the momentum of the bullet and block, Pf = (m+M)*v

where, v= initial speed of the (bullet+block)combination just after collision

By, conservation of momentum,

pi = pf

So, m*vi = (m+M)*v

So, v = m*vi/(m+M)

Now, for the combination to rise a height h = 0.18m,

The initial Kinetic energy of the combination must be converted to gravitational Potential energy at height h

So, 0.5*(m+M)*v^2 = (m+M)*g*h

So, v = sqrt(2gh)

So, m*(vi)/(m+M) = sqrt(2gh)

So, vi = ((m+M)/m)*sqrt(2gh)

So, vi = ((0.0045+1.95)/0.0045)*sqrt(2*9.8*0.18)

= 815.8 m/s <------------answer

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