Respuesta :
Answer:
[tex](a)\;V=0.058\;m/s\\(b)\;v_{2f}=0.1002\;m/s[/tex]
Explanation:
Given,
Mass of ball [tex]m_{1}=0.4[/tex] kg
Speed of ball [tex]v_{1i}=11[/tex] m/s
Mass of persona [tex]m_{2}=75[/tex] kg
(a) The person is at rest initially. So, by the conservation of momentum
[tex]p_{i}=p_{f}\\m_{1}v{1i}+m_{2}v_{2i}=(m_{1}+m_{2})V\\0.4\times 11+75 \times0=(0.4+75)V\\V=0.058\;m/s[/tex]
(b) Final speed of ball [tex]v_{1f}=-7.8\;m/s[/tex]
[tex]p_{i}=p_{f}\\m_{1}v{1i}+m_{2}v_{2i}=m_{1}v{1f}+m_{2}v_{2f}\\0.4\times 11+75 \times0=0.4\times(-7.8)+75v_{2f}\\v_{2f}=0.1002\;m/s[/tex]
Answer:
a) [tex]v=0.05835\ m.s^{-1}[/tex]
b) [tex]v=0.0171\ m.s^{-1}[/tex]
Explanation:
Given:
- initial speed of the ball before coming in contact to the catcher's body, [tex]u=11\ m.s^{-1}[/tex]
- mass of the ball, [tex]m=0.4\ kg[/tex]
- mass of the person catching the ball, [tex]m'=75\ kg[/tex]
a)
Since the person catching the ball is on a smooth surface, it will not dissipate the energy in the form of frictional heat.
So, when the person catches the ball in motion:
According to the conservation of momentum,
[tex]m.u=(m+m').v[/tex]
[tex]0.4\times 11=(0.4+75)\times v[/tex]
[tex]v=0.05835\ m.s^{-1}[/tex]
b)
When the ball hits and bounces off with a final velocity of 7.8 meters per second.
Then according to the conservation of linear momentum,
[tex]m.u=m.u'+m'.v[/tex]
where:
u' = final velocity of the ball after bounce-off
[tex]0.4\times 11=0.4\times 7.8+75\times v[/tex]
[tex]v=0.0171\ m.s^{-1}[/tex]
