If the 140 {\rm g} ball is moving horizontally at 26 {\rm m/s}, and the catch is made when the ballplayer is at the highest point of his leap, what is his speed immediately after stopping the ball?
To find the speed of the ballplayer immediately after stopping the ball, we can make use of the principle of conservation of momentum.
The formula for momentum is:
momentum (p) = mass (m) × velocity (v)
Since the ballplayer catches the ball, the momentum of the ball before catching should be equal to the momentum of the ballplayer after catching. Therefore, we can write:
momentum of ball before catching = momentum of ballplayer after catching
Now, let's calculate the momentum of the ball before catching. Given the mass of the ball (m) is 140 g, we need to convert the mass to kilograms:
mass of ball (m) = 140 g = 0.14 kg (divide by 1000 to convert grams to kilograms)
The velocity of the ball (v) is given as 26 m/s.
Now we can calculate the momentum of the ball before catching:
momentum of ball before catching = mass of ball × velocity of ball
= 0.14 kg × 26 m/s
Now we know the momentum of the ball before catching. According to the principle of conservation of momentum, this should be equal to the momentum of the ballplayer after catching. Let's denote the mass of the ballplayer as M and the velocity of the ballplayer as V.
momentum of ball before catching = momentum of ballplayer after catching
(0.14 kg × 26 m/s) = M × V
Since the ballplayer is stationary after catching the ball, the velocity of the ballplayer (V) is 0 m/s.
Therefore, the equation becomes:
0.14 kg × 26 m/s = M × 0 m/s
Simplifying the equation:
0.14 kg × 26 m/s = 0
Hence, the mass of the ballplayer (M) does not affect the result. We can't find the exact speed of the ballplayer immediately after stopping the ball.