Respuesta :
Answer:
a) [tex]u=0.05602\ m.s^{-1}[/tex]
b) [tex]u'=0.0169\ m.s^{-1}[/tex]
Explanation:
Given:
mass of the ball, [tex]m=0.4\ kg[/tex]
horizontal velocity of the ball, [tex]u=10\ m.s^{-1}[/tex]
mass of the person, [tex]M=71\ kg[/tex]
a)
Using the law of conservation of momentum:
[tex]m.v=(M+m).u[/tex]
[tex]0.4\times 10=(71+0.4)\times u[/tex]
[tex]u=0.05602\ m.s^{-1}[/tex]
b)
Given:
- rebound velocity of the ball, [tex]v'=7\ m.s^{-1}[/tex]
Using conservation of momentum,
[tex]m.v+M.u=m.v'+M.u'[/tex]
[tex]0.4\times10+71\times 0=0.4\times 7+71\times u'[/tex]
[tex]u'=0.0169\ m.s^{-1}[/tex]
Answer:
Explanation:
mass of ball, m = 0.4 kg
initial velocity of ball, u = 10 m/s
your mass, M = 71 kg
(a) let your speed after catching the ball is v.
Use conservation of momentum
m x u = (m + M) x v
0.4 x 10 = (0.4 + 71) x v
4 = 71.4 v
v = 0.056 m/s
(b) use conservation of momentum
(M + m) x v = m x u' + M x v'
(71 + 0.4) x 0.056 = 0.4 x 7 + 71 x v'
3.99 = 2.8 + 71 v'
v' = 0.017 m/s
