Question

A 0.049 kg bullet moving at 421 m/s strikes a stationary 4.7 kg wooden block. The bullet passes through the block and leaves with a velocity of 301 m/s. If the block was originally at rest, how fast does it move after being hit by the bullet? ANSWER ASAP

Answer 1

The block moves at 1.25 m/s

Step-by-step explanation:

Law Of Conservation Of Linear Momentum

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

`P=m_1v_1+m_2v_2`

If a collision occurs and the velocities change to v', the final momentum is:

`P'=m_1v'_1+m_2v'_2`

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

`m_1v_1+m_2v_2=m_1v'_1+m_2v'_2`

The bullet of m1=0.049 kg moves at v1=421 m/s. Then it strikes a wooden block of m2=4.7 kg originally at rest (v2=0).

After the collision, the bullet continues at v1'=301 m/s. The speed of the block can be calculated by solving for v2':

`\displaystyle v'_2=(m_1v_1+m_2v_2-m_1v'_1)/(m_2)`

`\displaystyle v'_2=(0.049*421+0-0.049*301)/(4.7)`

Calculating:

`\displaystyle v'_2=(5.88)/(4.7)=1.25\ m/s`

The block moves at 1.25 m/s