What is the impulse on the ball from the wall?

[A_lilah]

Homework Statement

a 250 g ball with a speed v of 6.6 m/s strikes a wall at an angle θ (the angle between the wall and the path of motion) of 30° and then rebounds with the same speed and angle. It is in contact with the wall for 6 ms. What is the impulse on the ball from the wall?

Homework Equations

J is the impulse
J = mvfinal - mvinitial
the change in momentum (P) = Jnet
P = mv

The Attempt at a Solution

First I broke the velocity vectors into components:

6.6m/s * cos(30) = 5.716m/s
6.6m/s * sin(30) = 3.3 m/s, so

Vo = 3.3i + 5.716j (m/s)
and
Vfinal = 3.3i - 5.716j (m/s)

and the momentum of each is the mass of the object * the velocity:

(.25kg)Vo = Po = .825i + 1.428j
(.25kg)vfinal = Pfinal = .825i - 1.428j

so Jnet = Pfinal - Pinitial = (.825-.825) i + (-1.428 -1.428)j
so Jnet = -2.856 in the j direction
so the answer should be -2.856, but it is not (and it is not 2.856 either)...

any help is appreciated.
Thanks!

 

[physixguru]

Have a re-look at your v(initial).

 

[Moetasim]

Your approach to solving this problem is correct, but there are a few errors in your calculations. The impulse on the ball from the wall can be calculated using the formula J = m(vfinal - vinitial). In this case, the initial velocity (vinitial) is the velocity of the ball before it hits the wall, which is 6.6 m/s at an angle of 30°. The final velocity (vfinal) is the velocity of the ball after it rebounds off the wall, which is also 6.6 m/s at an angle of 30°.

To calculate the components of the velocities, you used the wrong angle in your calculations. The angle between the wall and the path of motion is not 30°, but rather 60° (since the angle of incidence is equal to the angle of reflection). This means that the x-component of the initial velocity should be 6.6 m/s * cos(60°) = 3.3 m/s, and the y-component should be 6.6 m/s * sin(60°) = 5.716 m/s.

Using these values, the initial momentum (Pinitial) is (.25 kg)(3.3 m/s)i + (.25 kg)(5.716 m/s)j = .825i + 1.429j. Similarly, the final momentum (Pfinal) is (.25 kg)(3.3 m/s)i - (.25 kg)(5.716 m/s)j = .825i - 1.429j.

Therefore, the impulse on the ball from the wall is J = Pfinal - Pinitial = (.825i - 1.429j) - (.825i + 1.429j) = -2.858j Ns. So your answer was very close, but you made a small error in your calculations. Keep up the good work!

 

[Moetasim]

I would like to provide a response to the above content by pointing out that the impulse on the ball from the wall can be calculated using the equation J = FΔt, where F is the force applied by the wall and Δt is the time of contact. In this case, we can use the relationship between impulse and change in momentum (J = ΔP) to solve for the impulse.

First, we need to determine the change in momentum of the ball during the contact with the wall. This can be calculated using the equation ΔP = mΔv, where m is the mass of the ball and Δv is the change in velocity. Since the ball rebounds with the same speed and angle, there is no change in magnitude of the velocity. However, there is a change in direction, which results in a change in momentum.

Using the given information, we can calculate the change in velocity as follows:

Δv = 2vfinal - vinitial
= 2(6.6 m/s)cos(30°)i - (6.6 m/s)sin(30°)j - (6.6 m/s)cos(30°)i + (6.6 m/s)sin(30°)j
= 13.2 m/si - 2(6.6 m/s)sin(30°)j
= 13.2 m/si - (13.2 m/s)j

Now, we can calculate the change in momentum:

ΔP = mΔv
= (0.25 kg)(13.2 m/si - (13.2 m/s)j)
= 3.3i - 3.3j (kg·m/s)

Finally, we can use the relationship between impulse and change in momentum to calculate the impulse on the ball from the wall:

J = ΔP
= 3.3i - 3.3j (N·s)

Therefore, the impulse on the ball from the wall is 3.3 N·s in the i direction and -3.3 N·s in the j direction. This means that the force applied by the wall on the ball is in the opposite direction of the ball's initial velocity, which is expected since the ball is rebounding off the wall.