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irina [24]
3 years ago
12

A pitcher throws a 0.143-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just

before it makes contact with the bat. The batter then hits the ball straight back at the pitcher with a speed of 49 m/s. Assume the ball travels along the same line leaving the bat as it followed before contacting the bat. If the ball is in contact with the bat for 0.005 0 s, what is the magnitude of the average force exerted by the bat on the ball?
Physics
1 answer:
o-na [289]3 years ago
5 0

Answer:

200N appox

Explanation:

Step one:

Given data

mass of ball= 0.143kg

initial velocity of ball u= 42m/s

final velocity of the ball= 49m/s

time of impact= 0.005s

Step two

From the relation Ft=mΔv

we can find the average force as

F=mΔv/t

substitute

F=0.143*(49-42)/0.005

F=0.143*7/0.005

F=1.001/0.005

F=200.2

F= 200N appox

The average force is  200N

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Answer:

A) 112 m. B) 27.2 m C) 41.1 m/s i + 13.4 m/s j  D) 43.2 m/s

Explanation:

A) Once fired, no external forces act on the projectile in the x-direction, so it keeps moving to the right at constant speed, which is the projection on the x-axis of  the initial velocity vector:

v₀ₓ = v₀* cos 33º = 49 m/s* cos 33º = 41.1 m/s

In the y-direction, the component of the velocity can be found as the projection of v₀ on the y-axis, as follows:

v₀y = v₀* sin 33º = 49 m/s* sin 33º = 26.7 m/s

Both velocities are independent each other, as no one has a projection on the other.

In the vertical direction, the  projectile is in free fall all time, under the influence of gravity , which accelerates it downward.

So, at any time, in the vertical direction, the velocity can be calculated as follows:

vfy = v₀y -g*t (same equation as for an object thrown upwards)

When the object is at its maximum height, the velocity, in the vertical direction, will be momentarily zero, so we can find the time when this happens as follows:

vfy= 0 ⇒ v₀y = g*t ⇒ t = v₀y / g = 26.7 m/s / 9.8 m/s² = 2.72 s

As the time is the same for both movements, we can replace this value in the expression for the displacement x at constant speed, as follows:

x = v₀ₓ* t = 41.1 m/s* 2.72 s = 112 m

B) Like above, as the time is the same for both movements, we can find the time for the instant that the projectile hit the wall, as follows:

x = v₀ₓ* t ⇒ 55. 8 m = 41.1 m/s * t

⇒ t = 55. 8 m / 41.1 m/s = 1.36 s

We can replace this value of t in the equation for the vertical displacement, as follows:

Δy = v₀y*t -1/2*g*t² = (26.7m/s*1.36s) - 1/2*9.8m/s²*(1.36s)² = 27.2 m

C) The velocity of the projectile, at any time, has two components, one horizontal and one vertical.

As explained above, x-component is constant, equal to v₀x:

vx = v₀x i = 41.1 m/s i

For vy, we can apply acceleration definition, using the value of v₀y and t that we have just found:

vfy = voy - g*t = 26.7 m/s - 9.8m/s*1.36 sec = 13.4 m/s

vfy = 13.4 m/s j

v = 41.1 m/s i + 13.4 m/s j

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v = √(41.1 m/s)² +(13.4 m/s)² =43.2 m/s

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