Helpppppppppppppppp

Question

A 0.170-kg baseball traveling 34.0 m/s strikes the catcher's mitt, which, in bringing the ball to rest, recoils backward 15.0 cm .

Part A What was the magnitude of the average force applied by the ball on the glove?

Answer 1

The problem asks for the average force on the glove, which in a direct calculation would require knowledge about the mass of the glove and the acceleration of the glove. But no information about the glove is given. By Newton’s 3rd law, the force exerted by the ball on the glove is equal and opposite to the force exerted by the glove on the ball. So calculate the average force on the ball, and then take the opposite of that result to find the average force on the glove. The average force on the ball is its mass times its average acceleration. Find the acceleration of the ball, with

v = 0, v₀ = 34.0 m/s, and x - x₀ = 0.150 m. The initial direction of the ball is the positive direction.

a_avg = (v² - v_o²)/2(x - x₀) = 0 - (34.0 m/s)²/2(0.150 m) = - 3853.33 m/s²

F_avg = ma_avg = (0.170Kg)(-3853.33 m/s²) = 6.5507x10² N

Thus the average force on the glove was 655 N, in the direction of the initial velocity of the ball

Answer 2

The ball undergoes a negative acceleration when it hits the glove.

The negative acceleration can be found from V_final² = V_initial² + 2 x a x D where D is 0.15 m (15 cm).

V_initial equals 34.0 m/s, and V_final = 0 m/s.

Now Solve for a to get - 3.85 x 10³ m/s².

Knowing the acceleration, we can calculate the time from V_final - V_initial = a x t.

So t = 0 - 34.0 m/s / -3.85 x 10³ m/s² = 8.83 x 10^-3 s.

Average force is the change in momentum per unit time, so F_avg = (M_initial - M_final )/ time

F_avg = = 34.0 m/s x 0.170 kg - 0 / 8.83 x 10^-3 s = 655 N

Please check math.