Question

To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0146-kg bullet is fired straight up at a falling wooden block that has a mass of 2.55 kg. The bullet has a speed of 816 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. Number Units.

Answer 1

Answer:

0.25223 seconds.

Explanation:

$m_1$ = Mass of bullet = 0.0146 kg

$m_2$ = Mass of block = 2.55 kg

v = Combined velocity

$u_1$ = Velocity of bullet = 816 m/s

g = Acceleration due to gravity = 9.81 m/s²

As linear momentum is conserved

$m_1u_1 + m_2u_2 =(m_1 + m_2)v$

Now $u_2=v$ as the block (with the bullet in it) reverses direction and rises,

$m_1u_1 + m_2v =(m_1 + m_2)v\\\Rightarrow m_1u_1=(m_1 + m_2)v-m_2v\\\Rightarrow 0.0146\times 816=(0.0146 + 2.55)v-(-2.55v)\\\Rightarrow 11.9136=2.2646v+2.55v\\\Rightarrow 11.9136=4.8146v\\\Rightarrow v=\frac{11.9136}{4.8146}\\\Rightarrow v=2.47447\ m/s$

Equation of motion

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{2.47447-0}{9.81}\\\Rightarrow t=0.25223\ s$

The time t is 0.25223 seconds.