Respuesta :
Given data:
The mass of bowling ball is m=15 kg.
The height of ramp is h=5 m.
The velocity of bowling ball just before the collision can be calculated as,
[tex]\begin{gathered} v=\sqrt[]{2gh}\ldots\ldots\text{.}\mathrm{}(1) \\ v=\sqrt[]{2\times9.8\times5} \\ v=9.9\text{ m/s} \end{gathered}[/tex]Part (a)
The stationary mass is M=15 kg.
The velocity of bowling ball after collision will be,
[tex]\begin{gathered} v^{\prime}=\frac{m-M}{m+M}v \\ v^{\prime}=\frac{15-15}{15+15}(9.9) \\ v^{\prime}=0 \end{gathered}[/tex]Since the velocity of bowling ball after the collision is zero, therefore the ball will not move back.
Part (b)
The stationary mass is M=30 kg.
The velocity of bowling ball after collision will be,
[tex]\begin{gathered} v^{\prime}=\frac{m-M}{m+M}v \\ v^{\prime}=\frac{15-30}{15+30}(9.9) \\ v^{\prime}=-3.3\text{ m/s} \end{gathered}[/tex]Here, negative sign indicates that the ball is movin in opposite direction after the collision.
The velocity after the elastic collision is 3.3 m/s. So, the height back up by the ball can be calculated by substitute the value of velocity in equation (1),
[tex]\begin{gathered} v^{\prime}=\sqrt[]{2gh^{\prime}} \\ 3.3=\sqrt[]{2\times9.8\times h^{\prime}} \\ h^{\prime}=0.555\text{ m} \end{gathered}[/tex]Thus, the ball will returned to a height of 0.555 m.
Part (c)
The stationary mass is M=1000 kg.
The velocity of bowling ball after collision will be,
[tex]\begin{gathered} v^{\prime}=\frac{m-M}{m+M}v \\ v^{\prime}=\frac{15-1000}{15+1000}(9.9) \\ v^{\prime}=-9.6\text{ m/s} \end{gathered}[/tex]The velocity after the elastic collision is 9.6 m/s.
From equation (1),
[tex]\begin{gathered} v^{\prime}=\sqrt[]{2gh^{\prime}} \\ 9.6=\sqrt[]{2\times9.8\times h^{\prime}} \\ h^{\prime}=4.7\text{ m} \end{gathered}[/tex]Thus, the ball will returned to a height of 4.7 m.
