Ar whar speed must a 1s0-ig fooball lav oving so ave the sae oment bullet avelin

Question

Ar whar speed must a 1s0-ig fooball lav oving so ave the sae oment bullet aveling at 300mS as a 15.0- 8. 5. A 0.140-kg baseball is dropped and reaches a speed of 1.20 m/s just before it hits the ground.I rebounds with a speed of 1.00 m/s. What is the change of the ball's momentum? 9. A batter hits a 0.140-kg baseball that was approadhing him at 400 m/s and, as a result, the leaves the bat at 30.0 m/s in the direction of the pitcher. What is the magnitude of the impu leaves the bat at 30.0 m/s in the direction of the pitcher. What is the magnitude of the delivered to the baseball?

Explanation / Answer

here,
Ques 7 :
mass of player, mp = 150 kg

mass of bullet, mf = 15 g = 0.015 kg
velocity of bullet, v = 300 m/s

From Conservation of Momentum :
mp*vp = mf*v

Solving for velocity of player, vp = mf*v/mp
vp = 0.015 * 300/150
vp = 0.03 kg.m/s

Ques 8:
Assuming Downwards direction to be positive,
Change in moemntum , m = m(vf -vi)

(vf si final velocity, vi is initial velocity)
P = 0.140(-1 - 1.20)
P = -0.308 kg.m/s

-ve sign implise simply direction of net momentum

Ques :9

Assuming towards direction to be positive

mass of batter, m = 0.140 kg

Impulse = Chnage in momentum
impulse = m(vf - vi)
impulse = 0.140(-30 - 40)
impulse = -9.8 kg.m/s

-ve sign implise simply direction of net impulse

Ques 10:
mass skater 1, m1 = 45 kg
velocity of skater 1, v1 = 0.375 m/s

mass of skater 2, m2 = 60 kg

from conservation of momemtum,
m1v1 = m2*v2

solving for velocity of skater 2, v2 = m1*v1/m2
v2 = 45 * 0.375/60
v2 = 0.281 m/s

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