Question

To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.00970-kg bullet is fired straight up at a falling wooden block that has a mass of 2.78 kg. The bullet has a speed of 733 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occurs. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t.

Answer 1

For the block

Let, M = mass of the wooden block

m = mass of the bullet

u = initial velocity of the block

v = final velocity of the block

g = acceleration due to gravity

t = time of fall before collision

Since,

v = u + gt

v = gt ............eq(1)

( As the initial velocity u of the block is zero)

Distance covered by the wooden block while falling for time t

H = ut + (1/2)gt²

H = (1/2)gt²

(As the initial velocity u is zero)

At the time of collision wooden block has velocity v = gt in the downward direction whereas the bullet has velocity 733m/s in the upward direction.

Let v₀ be the combined velocity of the block and the bullet after collision

Using conservation of linear momentum

m$\times$733 - Mv = (m+M)v₀

From eq(1) , using v = gt

v₀ = (733m - Mgt) / (m+M) ...........eq(3)

After collision the wooden block + bullet system rises to height H (top of the building)

Using conservation of energy

(M+m)gH = (1/2)$\times$(m+M)v₀²

gH = (1/2)v₀²

Putting the value of H from eq(2) and the value of v₀ from eq(3)

g$\times$(1/2)gt² = (1/2)$\times$ { (733m - Mgt) / (m + M) }²

(gt)² = { (733m - Mgt) / (m+M) }²

gt = (733m - Mgt) / (m+M)

(M+m)gt = 733m - Mgt

(2M+m)gt = 733m

t = 733m / (2M+m)g

On putting, m = 0.0097 kg

M = 2.78 kg

g = 9.8 m/s²

t = 0.13 s