A 2.5kg ball strikes a wall with a velocity of 8.5m/s to the left. The ball bounces off the wall with a velocity of 7.5m/s to the right. If the ball is in contact with the wall for .25s, what is the constant force exerted on the ball by the wall?
calculate change in the momentum. remember that after it experiences this "perfectly elastic collision" it has a negative velocity. Therefore the change in momentum would be p = abs|m1v1 + m2v2|
change in momentum p = Ft
Force would be the average force or in this case the constant force
To find the constant force exerted on the ball by the wall, we can use Newton's second law of motion which states that force (F) is equal to the product of mass (m) and acceleration (a).
First, let's calculate the change in velocity of the ball during the collision with the wall. The initial velocity of the ball is 8.5 m/s to the left, and it bounces off the wall with a velocity of 7.5 m/s to the right. Therefore, we have a total change in velocity of:
Change in velocity = final velocity - initial velocity
= 7.5 m/s - (-8.5 m/s)
= 16 m/s
Next, we need to find the acceleration of the ball during the collision. We can use the equation:
Acceleration = Change in velocity / Time
Given that the ball is in contact with the wall for 0.25 seconds, we have:
Acceleration = 16 m/s / 0.25 s
= 64 m/s^2
Now that we have the acceleration, we can find the force exerted on the ball using Newton's second law:
Force (F) = mass (m) * acceleration (a)
Given that the mass of the ball is 2.5 kg, we have:
Force (F) = 2.5 kg * 64 m/s^2
= 160 N
Therefore, the constant force exerted on the ball by the wall is 160 Newtons.
To find the constant force exerted on the ball by the wall, we can use Newton's second law of motion, which states that the force exerted on an object is equal to the rate of change of its momentum. The formula for linear momentum is given by:
Momentum (p) = mass (m) x velocity (v)
We can calculate the initial momentum (p_initial) of the ball before it hits the wall using the given values of mass (m = 2.5 kg) and velocity (v_initial = 8.5 m/s):
p_initial = m x v_initial
= 2.5 kg x 8.5 m/s
= 21.25 kg·m/s
Next, we can calculate the final momentum (p_final) of the ball after it bounces off the wall using the given value of velocity (v_final = 7.5 m/s):
p_final = m x v_final
= 2.5 kg x 7.5 m/s
= 18.75 kg·m/s
The change in momentum (Δp) of the ball is given by the difference between the final and initial momentum:
Δp = p_final - p_initial
= 18.75 kg·m/s - 21.25 kg·m/s
= -2.5 kg·m/s (negative sign indicates a change in direction)
Now, we can calculate the average force (F_average) exerted on the ball during the contact time (t = 0.25 s) using the formula:
F_average = Δp / t
Substituting the values:
F_average = -2.5 kg·m/s / 0.25 s
= -10 kg·m/s² (negative sign indicates direction opposite to velocity)
Therefore, the constant force exerted on the ball by the wall is 10 N to the right.