Help with physics question asap!!!?

Question

A ball of mass 10.3 g with a speed of 20.4 m/s strikes a wall at an angle 17.0o and then rebounds with the same speed and angle. It is in contact with the wall for 42.0 ms. What is the magnitude of the impulse associated with the collision force?

the wall is vertical if it makes a difference

Answer 1

The impulse-momentum relationship is F*deltat = m*deltaV

m = 13g = .013kg
v = 20.4m/s
t = 42ms = .042s

I don't think the angle of incidence matters since it bounces off at the same angle.

Re-arranging the equation, F = mv/t

F = [(.013)(20.4)] / (.042)
=6.31N

But I guess since impulse = F*delta t, 6.31*.042 = .265 Newton-seconds.

Answer 2

The impulse only affects the horizontal component of the velocity (because the wall is assumed to exert only normal forces).

Write the component vectors of the ball's velocity both before and after the collision - the component vector which is parallel to the wall has not changed but the horizontal component has switched directions. It is the impulse from the wall that has caused the change in the normal velocity vector (and a change in vector velocity is an acceleration, which must come from a force....)

But you must know the angle of attack in order to write down the component vectors of the velocity; otherwise you only know the total speed.

Answer 3

Principle: Impulse= change in momentum
Since momentum=mv, it is a vector quantity.
The momentum is 10.3 gx20.4 m/s=202 g-m/sec.
If momentum vector is converted to its x and y components, the "incoming" momentum component of 202 Sin17 deg. becomes an outgoing component of 202 Sin 17 deg in the opposite direction. The other component continues in the same direction. So the change in momentum is 404 Sin 17 deg g-m/sec = F t
404 Sin17 deg=F (.0042) sec, F=10^5 Sin17 dynes (appx)