Question

A bat hits a moving baseball. If the bat delivers a net eastward impulse of 0.80 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)? (NEVER include units in the answer to a numerical question.)

Answer 1

By applying the principle of impulse and change in momentum, we calculate the mass of the baseball to be 84.21 grams using the given impulse and the change in velocity from the initial to the final state.

Step-by-step explanation:

To calculate the mass of the baseball, we can use the concept of impulse and the change in momentum. The impulse on an object is equal to the change in momentum it experiences, which is the product of the mass (m) of the object and its velocity change (Δv). The impulse (J) applied to the baseball by the bat can be calculated using the formula:

J = m Δv

This can be rearranged to solve for m as follows:

m = J / Δv

Given that the net eastward impulse is 0.80 N-s and the initial velocity (vi) is 3.8 m/s west (which we consider negative) and the final velocity (vf) is 5.7 m/s east (positive), the change in velocity (Δv) is the final velocity minus the initial velocity:

Δv = vf - (-vi)

Δv = 5.7 m/s - (-3.8 m/s)

Δv = 9.5 m/s

Now we can use this value to find the mass of the baseball:

m = 0.80 N-s / 9.5 m/s

m = 0.08421 kg

To convert the mass of the ball into grams, we multiply by 1000, since there are 1000 grams in a kilogram:

m = 0.08421 kg × 1000 g/kg

m = 84.21 grams

Therefore, the mass of the baseball is 84.21 grams.

Answer 2

0.084 kg

Step-by-step explanation:

I = 0.80 N-s (East wards) = 0.80 i N-s

u = 3.8 m/s = - 3.8 i m/s

v = 5.7 m/s = 5.7 i m/s

Let m be the mass of bat.

I = m (v - u)

0.8 i = m ( 5.7 i + 3.8 i)

0.8 i = m x 9.5 i

m = 0.084 kg