Question

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at and it leaves the bat traveling to the left at an angle of above horizontal with a speed of If the ball and bat are in contact for 1.75 ms, find the horizontal and vertical components of the average force on the ball.

Answer 1

Answer

given,

mass of the baseball = 0.145 Kg

Assuming the horizontal velocity of ball = V x = 50 m/s

ball left at an angle of = 30°

At the speed of 65 m/s

time of contact  = 1.75 ms

velocity of time along horizontal direction

v_{horizontal} = -65 cos 30° - 50

                        = -106.29 m/s

Impulse = force x time

impulse is equal to change in momentum

now,

force x time = m v

$F = \dfrac{mv}{t}$

$F = \dfrac{0.145 \times (-106.29)}{1.75 \times 10^{-3}}$

F_{horizontal} = - 8.806 kN

now velocity in vertical component

v_{vertical} = 65 sin 30°

                  = 32.5 m/s

$F = \dfrac{mv}{t}$

$F = \dfrac{0.145 \times (32.5)}{1.75 \times 10^{-3}}$

F_{vertical} = 2.692 kN

Answer 2

(a) The horizontal component of the average force on the ball is 7,633.63 N.

(b) The vertical component of the average force on the ball is 2,929 N.

The given parameters;

• mass of the bat, m₁ = 0.145 kg
• time, t = 1.75 ms
• initial velocity of the ball to right, v₀ = 50 m/s
• final velocity of the ball to left v = 55 m/s  at 40⁰

The horizontal component of the ball's acceleration is calculated as follows;

$-v_f_x = v_0_x - a_xt\\\\a_xt = v_0_x + v_f_x\\\\a_x = \frac{v_0_x + v_f_x}{t} \\\\a_x = \frac{50 \ + \ 55cos(40)}{0.00175} \\\\a_x = 52,645.7 \ m/s^2$

The horizontal component of the average force on the ball is calculated as follows;

$F_x = ma_x\\\\F_x = 0.145 \times 52,645.7 \\\\F_x = 7,633.63 \ N$

The vertical component of the ball's acceleration is calculated as follows;

$v_f_y = v_0y + a_yt\\\\v_f_y = 0+ a_yt\\\\a_y = \frac{v_f_y}{t} \\\\a_y = \frac{55 \times sin(40)}{0.00175} \\\\a_y = 20,200 \ m/s^2$

The vertical component of the average force on the ball is calculated as follows;

$F_y = ma_y\\\\F_y = 0.145 \times 20,200\\\\F_y = 2,929 \ N$

"Your question is not complete, it seems to be missing the following information";

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 50.0 m/s , and it leaves the bat traveling to the left at an angle of 40 ∘ above horizontal with a speed of 55.0 m/s . The ball and bat are in contact for 1.75 ms .

find the horizontal and vertical components of the average force on the ball.

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