How Is Impulse Calculated in a Two-Dimensional Collision?

[chaotixmonjuish]

A ball of mass 7.0 g with a speed of 25.2 m/s strikes a wall at an angle 23.0 ° and then rebounds with the same speed and angle. It is in contact with the wall for 39.0 ms. What is the magnitude of the impulse associated with the collision force?
What is the average force exerted by the ball on the wall?

The only equation I know for impulse is:
F dt=m*v

I'm not sure how to treat this problem since it has an angle, and impulse isnew.

 

[liskawc]

try drawin a picture : then set some coordinates and then use trigonometric functions to see what's the speed in the axis that's in the right angle towards your wall (sounds confusing?) and then just apply it to the equation you have

 

[ogb p]

I would approach this problem by first breaking it down into its components. The impulse, which is the change in momentum, can be calculated using the equation F*dt=mvf-mvi, where F is the force, dt is the time interval, mvf is the final momentum of the ball, and mvi is the initial momentum of the ball.

In this case, we can calculate the initial and final momentums of the ball using the given mass and velocity. The initial momentum would be 7.0 g * 25.2 m/s = 176.4 g*m/s. To calculate the final momentum, we need to take into account the angle at which the ball rebounds. Using trigonometry, we can determine that the horizontal component of the final momentum would be 7.0 g * 25.2 m/s * cos(23.0°) = 165.1 g*m/s, and the vertical component would be 7.0 g * 25.2 m/s * sin(23.0°) = 66.2 g*m/s.

Next, we can calculate the magnitude of the impulse by subtracting the initial momentum from the final momentum. This would be 165.1 g*m/s - 176.4 g*m/s = -11.3 g*m/s. Since impulse is a vector quantity, the negative sign indicates that the direction of the impulse is opposite to the initial momentum of the ball.

To calculate the average force exerted by the ball on the wall, we can use the equation F=mvf/dt. Since we know the final momentum and the time interval of contact with the wall, we can plug in the values and get F=165.1 g*m/s / 0.039 s = 4230 g*m/s^2. This is equivalent to 4.23 N, which is the average force exerted by the ball on the wall during the 39.0 ms of contact.

In summary, the magnitude of the impulse associated with the collision force is 11.3 g*m/s and the average force exerted by the ball on the wall is 4.23 N. It is important to note that these calculations assume an idealized scenario and may differ slightly from real-world situations due to factors such as air resistance and imperfections in the ball and wall surfaces.