# Physics - Newton's Second Law Q?

A 0.100 kg baseball travelling 30.0 m/s strikes the catcher’s mitt which, in bringing the ball to rest, recoils backwards 10.0 cm. What was the average force applied by the ball on the glove?

### 3 Answers

- silver dLv 77 years agoFavorite Answer
Using Newton’s 2nd law is the long way of solving this problem, but here goes.

First, we have to find the deceleration of the ball, coming to rest in the catcher’s mitt. The initial velocity of the baseball is 30 m/S, and the final velocity of the ball is 0.0 m/s; Vave = [V0 + Vf ]/2 = 15.0 m/s. During its deceleration, the ball has an average velocity of 15 m/S. We need this information to determine how long the deceleration takes. Remember: Speed = distance (Δ x) / time (seconds). We know that Δ x = 10 cm (the distance the mitt moves), and the average velocity of the ball during its deceleration = 15 m/S. We can rearrange the equation to solve for time: t=d/S. 0.1m/15.0 m/S = 6.67mS; so the ball takes 6.67 milliseconds to totally decelerate to 0.0 m/S. Acceleration = Δ V / Δ t. Δ V = -30.0 m/S, and t = 6.67 mS. (-30.0 m/s) / 6.67 mS = -4500 m/S^2. We know the mass of the ball is 0.1 kg. Plugging in the numbers to the famous equation F=ma, we get 0.1 kg * -4500 m/S^2= -450 Newtons of force. <-- you can disregard the negative sign; whether it’s positive or negative depends on your frame of reference.

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- Anonymous7 years ago
You can use the conservation of energy.

The energy in the ball is used to move the glove/catcher 10.0cm

Energy = 1/2 m * v^2 = Work = W = F_avg * d

F_avg = 1/2 m * v^2 / d = 1/2 * (0.100 kg) * ( 30.0 m/s)^2 / 10.0 cm

F_avg = 1/2 m * v^2 / d = 1/2 * (0.100 kg) * ( 30.0 m/s)^2 / 0.1 m

F_avg = 1/2 * 900 = 450 kg m /s^2 = 450 N

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- Randy PLv 77 years ago
The work done by the glove F * d is equal to the initial kinetic energy (1/2)mv^2 of the ball.

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