How Is Impulse Calculated in Tennis Ball Physics?

[Northbysouth]

Homework Statement

To warm up for a match, a tennis player hits the 58.0g ball vertically with her racket.
If the ball is stationary just before it is hit and goes 5.50m high, what impulse did she impart to it?

I know that the answer is 0.602 kgm/s but I don't understand why this is the answer

Homework Equations

P=mv

p_total = P_initial + P_final

The Attempt at a Solution

P_initial = (0.058kg)(0 m/s)
= 0

P_final = (0.058kg)(v_final

I'm not sure where to go from here.
I think that k_initial+U_initial = k_final+u_final has something to do with it
But I'm not sure.

 

[cepheid]

Energy is conserved. If the ball reaches a height of 5.50 m, it has gained a certain amount of gravitational potential energy (you can calculate the amount). That energy had to come from somewhere. From this fact, you can deduce the speed of the ball at the end of the impact with the racket. ;)

 

[fredb]

compute v(0) tennis ball: v(t)[/2] = v(0)[/2] - 2*g*d → 0 = v(0)[/2] - 2*9.8*5.5 = 10.38 m/s.

ƩP(initial) = ƩP(final) (just before and just after the hit). In general p=mv.

Initial m(b)*v(b) + m(r)*v(r) = 0+ m(r)*v(r)

Final m(b)*v(b) + m(r)*v(r) = 0.058*10.38=0.602 kgm/s

So initial momentum racket is 0,602 kgm/s.

 

[cepheid]

Fredb, we do NOT provide complete solutions to homework problems on this site. It is against site rules.

 

[Nickie]

I would first clarify the given information and equations to make sure we are on the same page. The given equation for impulse is actually P= FΔt, where P is impulse, F is force, and Δt is the time interval over which the force is applied. This equation can also be written as P= mv, where m is mass and v is velocity.

In the problem, we are given the mass of the ball (58.0g or 0.058kg), the height it reaches (5.50m), and the fact that it starts from rest (stationary). We can use the equation for potential energy (U=mgh) to find the initial potential energy of the ball before it is hit, which is equal to the final potential energy when it reaches the maximum height. So we have:

Uinitial = Ufinal
mgh = mgh
(0.058kg)(9.8m/s^2)(0m) = (0.058kg)(9.8m/s^2)(5.50m)
0 = 0.320 kgm^2/s^2

This means that the initial and final kinetic energies must also be equal, so we can use the equation for kinetic energy (K= 1/2mv^2) to find the final velocity of the ball when it reaches the maximum height. We have:

Kinitial = Kfinal
0 = (0.5)(0.058kg)vfinal^2
0 = 0.029kgvfinal^2
vfinal = 0 m/s

Since the ball is at rest at the maximum height, its final velocity is 0 m/s.

Now, we can use the equation for impulse (P= mv) to find the impulse imparted to the ball by the tennis player:

P= (0.058kg)(0m/s)
= 0 kgm/s
= 0.602 kgm/s (rounded to three significant figures)

This means that the tennis player imparted an impulse of 0.602 kgm/s to the ball when she hit it vertically with her racket. This impulse is equal to the change in momentum of the ball, which is why it is also equal to the final momentum (0 kgm/s) minus the initial momentum (0 kgm/s). I hope this explanation helps to clarify the solution to the problem.