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 ball of mass 0.400 kg that is traveling horizontally at 11.0 m/s. Your mass is 75.0 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 that afterwards it is moving horizontally at 7.80m/s in the opposite direction, what is your speed after the collision?

Answer 1

Answer:

a) $v=0.05835\ m.s^{-1}$

b) $v=0.0171\ m.s^{-1}$

Explanation:

Given:

• initial speed of the ball before coming in contact to the catcher's body, $u=11\ m.s^{-1}$

• mass of the ball, $m=0.4\ kg$

• mass of the person catching the ball, $m'=75\ kg$

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,

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

$0.4\times 11=(0.4+75)\times v$

$v=0.05835\ m.s^{-1}$

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,

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

where:

u' = final velocity of the ball after bounce-off

$0.4\times 11=0.4\times 7.8+75\times v$

$v=0.0171\ m.s^{-1}$

Answer 2

Answer:

$(a)\;V=0.058\;m/s\\(b)\;v_{2f}=0.1002\;m/s$

Explanation:

Given,

Mass of ball $m_{1}=0.4$ kg

Speed of ball $v_{1i}=11$ m/s

Mass of persona $m_{2}=75$ kg

(a) The person is at rest initially. So, by the conservation of momentum

$p_{i}=p_{f}\\m_{1}v{1i}+m_{2}v_{2i}=(m_{1}+m_{2})V\\0.4\times 11+75 \times0=(0.4+75)V\\V=0.058\;m/s$

(b) Final speed of ball $v_{1f}=-7.8\;m/s$

$p_{i}=p_{f}\\m_{1}v{1i}+m_{2}v_{2i}=m_{1}v{1f}+m_{2}v_{2f}\\0.4\times 11+75 \times0=0.4\times(-7.8)+75v_{2f}\\v_{2f}=0.1002\;m/s$