Respuesta :
Answer:
(a). The speed of the ball after collision is 2.01 m/s.
(b). The speed of the block after collision 1.11 m/s.
Explanation:
Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.
Given that,
Mass of steel block = 2.30 kg
Mass of ball = 0.500 kg
Length of cord = 50.0 cm
We need to calculate the initial speed of the ball
Using conservation of energy
[tex]\dfrac{1}{2}mv^2=mgl[/tex]
[tex]v=\sqrt{2gl}[/tex]
Put the value into the formula
[tex]u=\sqrt{2\times9.8\times50.0\times10^{-2}}[/tex]
[tex]u=3.13\ m/s[/tex]
The initial speed of the ball [tex]u_{1}=3.13\ m/s[/tex]
The initial speed of the block [tex]u_{2}=0[/tex]
(a). We need to calculate the speed of the ball after collision
Using formula of collision
[tex]v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}[/tex]
Put the value into the formula
[tex]v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13[/tex]
[tex]v_{1}=-2.01\ m/s[/tex]
Negative sign shows the opposite direction of initial direction.
(b). We need to calculate the speed of the block after collision
Using formula of collision
[tex]v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}[/tex]
Put the value into the formula
[tex]v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0[/tex]
[tex]v_{2}=1.11\ m/s[/tex]
Hence, (a). The speed of the ball after collision is 2.01 m/s.
(b). The speed of the block after collision 1.11 m/s.
