Terrie Lang

2022-01-10

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 0.600-kg ball that is traveling horizontally at 10.0 m/s. Your mass is 70.0 kg.

(a) If you catch the ball, with what speed do you and the ball move afterward?

(b) If the ball hits you and bounces off your chest, so afterward it is moving horizontally at 8.0 m/s in the opposite direction, what is your speed after the collision?

(a) If you catch the ball, with what speed do you and the ball move afterward?

(b) If the ball hits you and bounces off your chest, so afterward it is moving horizontally at 8.0 m/s in the opposite direction, what is your speed after the collision?

redhotdevil13l3

Beginner2022-01-11Added 30 answers

Step 1

a) Let the mass of the ball be$m=0.6kg$ and $M=70kg$ . The ball is initially traveling towards us with the speed of $v}_{0}=10\frac{m}{s$ . We catch the ball and are moving with the ball with speed v on the ice. Since there is no friction, there are no forces acting on us and the ball in the horizontal direction. Hence, the total horizontal momentum is conserved and we have

$m{v}_{0}=(m+M)v\Rightarrow v=\frac{m}{m+M}{v}_{0}=\frac{0.6}{70.6}10\frac{m}{s}=0.085\frac{m}{s}$

a) Let the mass of the ball be

Neil Dismukes

Beginner2022-01-12Added 37 answers

Step 2

b) Now, the speed of the ball after the collision is$v}_{b}=-8\frac{m}{s$ (the minus sign comes from the fact that it travels in the opposite direction after the hit) and we have to find our speed V. Again, from the law of conservation of momentum, we have

$m{v}_{0}=m{v}_{b}+MV\Rightarrow V=\frac{m}{M}({v}_{0}-{v}_{b})=\frac{0.6}{70}(10+8)\frac{m}{s}=0.154\frac{m}{s}$

b) Now, the speed of the ball after the collision is

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