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

m = 105.26 g

Step-by-step 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 *(5.7-(-3.8))`

`m = (1)/(9.5)`

m = 0.10526 Kg

m = 105.26 g

Mass of the ball is equal to 105.26 g.

Answer 2

105 g

Step-by-step 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=(1)/(9.5)=0.105 kg`

`m=0.105* 1000=105 g`

Where 1 kg=1000 g