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

Answer:

The block moves at 1.25 m/s

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':

$\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2}$

$\displaystyle v'_2=\frac{0.049*421+0-0.049*301}{4.7}$

Calculating:

$\displaystyle v'_2=\frac{5.88}{4.7}=1.25\ m/s$

The block moves at 1.25 m/s