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:
[tex]P=m_1v_1+m_2v_2[/tex]
If a collision occurs and the velocities change to v', the final momentum is:
[tex]P'=m_1v'_1+m_2v'_2[/tex]
Since the total momentum is conserved, then:
P = P'
Or, equivalently:
[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]
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':
[tex]\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2}[/tex]
[tex]\displaystyle v'_2=\frac{0.049*421+0-0.049*301}{4.7}[/tex]
Calculating:
[tex]\displaystyle v'_2=\frac{5.88}{4.7}=1.25\ m/s[/tex]
The block moves at 1.25 m/s
