Question

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a 0.400-kg ball that is traveling horizontally at 10.0 m/s. Your mass is 71.0 kg.
(a) If you catch the ball, with what speed do you and the ball move afterwards?
(b) If the ball hits you and bounces off your chest, so afterwards it is moving horizontally at 7.00 m/s in the opposite direction, what is your speed after the collision?

Answer 1

Answer:

a) $u=0.05602\ m.s^{-1}$

b) $u'=0.0169\ m.s^{-1}$

Explanation:

Given:

mass of the ball, $m=0.4\ kg$

horizontal velocity of the ball, $u=10\ m.s^{-1}$

mass of the person, $M=71\ kg$

a)

Using the law of  conservation of momentum:

$m.v=(M+m).u$

$0.4\times 10=(71+0.4)\times u$

$u=0.05602\ m.s^{-1}$

b)

Given:

• rebound velocity of the ball, $v'=7\ m.s^{-1}$

Using conservation of momentum,

$m.v+M.u=m.v'+M.u'$

$0.4\times10+71\times 0=0.4\times 7+71\times u'$

$u'=0.0169\ m.s^{-1}$

Answer 2

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