1. A 0.75kg ball traveling horizontally to the right at 20m/s. A player strikes the ball and causes it to move in the opposite direction with a velocity of 24m/s.
a. What impulse was delivered by the player to the ball?
b. If this took 0.3 seconds to occur, What force must be involved to change the momentum?
c. Calculate the acceleration rate of this object.
Thanks for the help.
change of momentum= m(deltaV)=m(20+24)
impulse= change of momentum
Force= impulse/time
F=ma solve for a.
a. To find the impulse delivered by the player to the ball, we can use the formula:
Impulse = Change in momentum
The change in momentum can be calculated by subtracting the final momentum from the initial momentum:
Initial momentum = mass × initial velocity
Final momentum = mass × final velocity
Plugging in the given values, we have:
Initial momentum = 0.75kg × 20m/s = 15 kg·m/s
Final momentum = 0.75kg × (-24m/s) = -18 kg·m/s
So the change in momentum is:
Change in momentum = Final momentum - Initial momentum
Change in momentum = -18 kg·m/s - 15 kg·m/s = -33 kg·m/s
Therefore, the impulse delivered by the player to the ball is -33 kg·m/s.
b. To find the force involved in changing the momentum, we can use the formula:
Impulse = Force × time
Given that the time taken is 0.3 seconds, we can rearrange the formula to solve for force:
Force = Impulse / time
Plugging in the values, we have:
Force = (-33 kg·m/s) / (0.3 s) = -110 N
Therefore, the force involved in changing the momentum is -110 N.
c. To calculate the acceleration rate of the object, we can use the formula:
Acceleration = Change in velocity / time
Given that the change in velocity is the final velocity minus the initial velocity, we have:
Change in velocity = final velocity - initial velocity
Change in velocity = (-24 m/s) - (20 m/s) = -44 m/s
Dividing the change in velocity by the time taken, we have:
Acceleration = (-44 m/s) / (0.3 s) = -146.67 m/s^2
Therefore, the acceleration rate of the object is -146.67 m/s^2.