While standing at home plate, a batter swung a 2.5kg bat at 2.5 m/s toward an incoming mass. The mass was a 0.4 kg blob of clay, thrown toward the batter at 26 m/s, which stuck the the bat in an inelastic collision. What velocity of the velocity of the bat and blob of clay (which are stuck together) after the collision? Help fast will give brainliest
Show work

Answers

Answer 1

Answer:

5.74 m/s

Explanation:

m1v1+m2v2=(m1+m2)*vf

2.5*2.5+0.4*26=(2.9)vf

16.65=2.9vf

vf=5.74 m/s


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A 0.144 kg baseball approaches a batter with a speed of . The batter lines the ball directly back to the pitcher with a speed of . Find the impulse exerted on the ball . If the bat and ball were in contact for 0.012 sec , find the average force exerted on the ball by the bat 20 m/s 30 m/s

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Complete question:

A 0.144 kg baseball approaches a batter with a speed of 20m/s . The batter lines the ball directly back to the pitcher with a speed of 30m/s . Find the impulse exerted on the ball . If the bat and ball were in contact for 0.012 sec , find the average force exerted on the ball by the bat

Answer:

(1) The impulse exerted by the ball is 7.2 kg.m/s

(2) The average force exerted on the ball by the bat is 600 N.

Explanation:

Given;

mass of the baseball, m = 0.144 kg

velocity of the baseball, v₁ = 20 m/s

velocity of the batter, v₂ = -30 m/s (opposite direction to the ball's speed)

The impulse exerted by the ball is calculated as follows;

J = ΔP = mv₁ - mv₂

ΔP = m(v₁ - v₂)

ΔP = 0.144 [20 - (-30)]

ΔP = 0.144 ( 20 + 30 )

ΔP = 0.144 (50)

ΔP = 7.2 kg.m/s

The average force exerted on the ball by the bat is calculated as;

[tex]F = \frac{\Delta P }{t} \\\\where; t \ is \ time \ of \ contact= 0.012 \ s\\\\F = \frac{7.2}{0.012} \\\\F = 600 \ N[/tex]

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