Question

g A bat hits a moving baseball. If the bat delivers a net eastward impulse of 1.0 N-s and the ball starts with an initial horizontal velocity of 3.8 m/s to the west and leaves with a 5.7 m/s velocity to the east, what is the mass of the ball (in grams)

Answer 1

Answer:

m = 105.26 g

Explanation:

Given,

Impuse = 1 Ns

initial horizontal velocity = 3.8 m/s

final horizontal velocity = 5.7 m/s

Mass of the ball = ?

we know impulse is equal to change in momentum

$I = m (v_f-v_i)$

$1 = m \times(5.7-(-3.8))$

$m = \dfrac{1}{9.5}$

m = 0.10526 Kg

m = 105.26 g

Mass of the ball is equal to 105.26 g.

Answer 2

Answer:

105 g

Explanation:

We are given that

Impulse=I=1 N-s

Initial velocity,u=-3.8 m/s

Final velocity,v=5.7 m/s

We have to find the mass of the ball.

Impulse=Change in momentum

I=m(v-u)

Using the formula

$1=m(5.7-(-3.8))$

$1=(5.7+3.8)m$

$1=9.5m$

$m=\frac{1}{9.5}=0.105 kg$

$m=0.105\times 1000=105 g$

Where 1 kg=1000 g