A 0.280-kg volleyball approaches a player horizontally with a speed of 15.0 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 22.0 m/s. a)what impulse is delivered to the ball by the player? b)If the player's fist is in contact with the ball for 0.060 s, find the magnitude of the average force exerted on the player's fist?
impulse=change momentum=mass(Vf-Vi)
= mass(22-(-15)
then, force=impulse/time
To solve this problem, we can use the impulse-momentum principle. The impulse delivered to an object is equal to its change in momentum. Impulse is defined as the product of the force applied to an object and the time interval over which it acts.
a) To find the impulse delivered to the ball, we need to determine its change in momentum. The momentum of an object is given by the product of its mass and velocity.
Given:
Mass of the ball (m1) = 0.280 kg
Initial velocity (v1) = 15.0 m/s
Final velocity (v2) = -22.0 m/s (opposite direction)
So, the change in momentum (Δp) of the ball is:
Δp = m1 * (v2 - v1)
= 0.280 kg * (-22.0 m/s - 15.0 m/s)
= 0.280 kg * (-37.0 m/s)
= -10.36 kg·m/s
Therefore, the impulse delivered to the ball by the player is -10.36 kg·m/s.
b) To find the magnitude of the average force exerted on the player's fist, we can use the equation:
Impulse (J) = Force (F) * Time (Δt)
Rearranging the equation, we have:
F = J / Δt
Given:
Impulse (J) = -10.36 kg·m/s (from part a)
Time (Δt) = 0.060 s
Plugging in the values, the magnitude of the average force exerted on the player's fist is:
F = (-10.36 kg·m/s) / (0.060 s)
= -172.67 N
The negative sign indicates that the direction of the average force is opposite to the direction of motion.
Therefore, the magnitude of the average force exerted on the player's fist is 172.67 Newtons (N).
To answer both questions, we need to apply the principles of impulse and force. Here's how you can find the answers:
a) To calculate the impulse delivered to the ball by the player, we can use the impulse-momentum principle. The impulse is defined as the change in momentum. The formula for impulse (J) is:
J = Δp = m * Δv
where:
J is the impulse,
m is the mass of the object, and
Δv is the change in velocity.
In this case, the mass of the volleyball is 0.280 kg. The change in velocity is the final velocity (22.0 m/s) minus the initial velocity (15.0 m/s):
Δv = 22.0 m/s - (-15.0 m/s) = 22.0 m/s + 15.0 m/s = 37.0 m/s
Now we can calculate the impulse:
J = 0.280 kg * 37.0 m/s = 10.36 kg⋅m/s
Therefore, the impulse delivered to the ball by the player is 10.36 kg⋅m/s.
b) To find the magnitude of the average force exerted on the player's fist, we can use the formula:
F = Δp / Δt
where:
F is the force,
Δp is the change in momentum, and
Δt is the time interval.
As we calculated earlier, the impulse delivered to the ball is 10.36 kg⋅m/s. The time interval (Δt) is given as 0.060 s.
Now we can calculate the force:
F = 10.36 kg⋅m/s / 0.060 s = 172.67 N
Therefore, the magnitude of the average force exerted on the player's fist is approximately 172.67 Newtons.