Answer:
(a) Velocity of target is 2.7 m/s.
(b) Mass of target is 0.807 kg.
Explanation:
Mass of ball [tex]m_{1}=0.22\;kg[/tex]
initial speed of the ball [tex]v_{1i}=6.3\;m/s[/tex]
Final speed of the ball [tex]v_{2f}=-3.6\;m/s[/tex]
Part (a) The second ball is initially at rest. So, by the conservation of momentum,
[tex]p_{i}=p_{f}\\m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{1f}+m_{2}v_{2f}\\0.22\times6.3+m_{2}\times 0=0.22\times(-3.6)+m_2v_{2f}\\m_2v_{2f}=2.178\:\:\:\:\;\;\;\;\;\;\;\;\;\; ...(1)[/tex]
Now, the velocity of approach is equal to the velocity of separation,
[tex]v_{1i}-v_{2i}=v_{2f}-v_{1f}\\6.3-0=v_{2f}-(-3.6)\\v_{2f}=2.7\;m/s[/tex]
Part (b): From equation (1),
[tex]m_2v_{2f}=2.178\\m_2=\frac{2.178}{2.7}\\m_2=0.806\;kg[/tex]
