A 0.278-kg volleyball approaches a player horizontally with a speed of 16.2 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 22.4 m/s.
(a) What impulse is delivered to the ball by the player? (Take the direction of final velocity to be the positive direction. Indicate the direction with the sign of your answer.)
kg · m/s
(b) If the player's fist is in contact with the ball for 0.0600 s, find the magnitude of the average force exerted on the player's fist.
N
(a) The impulse delivered to an object is defined as the change in momentum of the object. In this case, the initial momentum of the volleyball is given by the product of its mass and initial velocity, while the final momentum is given by the product of its mass and final velocity.
Initial momentum = mass x initial velocity = 0.278 kg x 16.2 m/s = 4.4556 kg·m/s
Final momentum = mass x final velocity = 0.278 kg x (-22.4 m/s) = -6.2272 kg·m/s
The change in momentum is given by the final momentum minus the initial momentum:
Change in momentum = Final momentum - Initial momentum
Change in momentum = (-6.2272 kg·m/s) - (4.4556 kg·m/s)
Change in momentum = -10.6828 kg·m/s
Therefore, the impulse delivered to the ball by the player is -10.6828 kg·m/s. The negative sign indicates that the impulse is in the opposite direction of the initial momentum.
(b) The average force exerted on an object is given by the impulse delivered to the object divided by the time interval over which the impulse is exerted.
Average force = Impulse / Time interval
In this case, the impulse delivered to the ball is -10.6828 kg·m/s and the time interval is 0.0600 s:
Average force = (-10.6828 kg·m/s) / (0.0600 s)
Average force = -178.0467 N
Therefore, the magnitude of the average force exerted on the player's fist is 178.0467 N. The negative sign indicates that the force is exerted in the opposite direction to the final velocity of the ball.
To solve this problem, we will use the impulse-momentum principle, which states that the change in momentum of an object is equal to the impulse delivered to it. The impulse is the product of the force applied to the object and the time interval over which the force acts.
(a) We can calculate the impulse delivered to the ball by using the equation:
Impulse = Change in momentum
The initial momentum of the ball is given by:
Initial momentum = mass × initial velocity
Final momentum of the ball is given by:
Final momentum = mass × final velocity
Since the direction of the final velocity is opposite to the initial velocity, we need to consider the signs of the velocities in the calculation.
Initial momentum = -0.278 kg × 16.2 m/s (-ve sign because it is opposite to the final direction)
Final momentum = 0.278 kg × 22.4 m/s (final direction is positive)
Now, we can calculate the impulse:
Impulse = Final momentum - Initial momentum
Impulse = (0.278 kg × 22.4 m/s) - (-0.278 kg × 16.2 m/s)
Simplifying the equation gives:
Impulse = 6.2432 kg·m/s - (-4.5176 kg·m/s)
Impulse = 10.7608 kg·m/s
Therefore, the impulse delivered to the ball by the player is 10.7608 kg·m/s.
(b) The average force exerted on the ball can be calculated using the equation:
Average force = Impulse / Time interval
We can substitute the impulse and time interval values into the equation:
Average force = 10.7608 kg·m/s / 0.0600 s
Calculating the average force gives:
Average force = 179.346 N
Therefore, the magnitude of the average force exerted on the player's fist is 179.346 N.
To answer both parts of the question, we need to use the principles of impulse and momentum.
(a) The impulse delivered to an object is equal to the change in momentum it experiences. Impulse (J) can be calculated using the formula:
J = Δp = mvf - mvi
Where J is the impulse in kg · m/s, Δp is the change in momentum, mvf is the final momentum, and mvi is the initial momentum.
In this case, the initial momentum of the volleyball is the product of its mass and initial velocity:
mvi = (0.278 kg) × (16.2 m/s) = 4.4976 kg · m/s (taking the positive direction as initial velocity)
The final momentum of the volleyball is the product of its mass and final velocity:
mvf = (0.278 kg) × (-22.4 m/s) = -6.2272 kg · m/s (taking the opposite direction as final velocity)
Therefore, the change in momentum is:
Δp = mvf - mvi = -6.2272 kg · m/s - 4.4976 kg · m/s = -10.7248 kg · m/s
So, the impulse delivered to the ball by the player is -10.7248 kg · m/s.
(b) The average force exerted on an object can be calculated using the formula:
F = Δp / Δt
Where F is the force in Newtons (N), Δp is the change in momentum, and Δt is the time interval over which the force is exerted.
In this case, the change in momentum is the same as from part (a):
Δp = -10.7248 kg · m/s
The time interval is given as 0.0600 s.
Therefore, the magnitude of the average force exerted on the player's fist is:
F = Δp / Δt = (-10.7248 kg · m/s) / (0.0600 s) = -178.747 N
Since force is a vector quantity, the negative sign indicates that the force is exerted in the opposite direction of the motion of the ball.
Force = change in momentum/time = m a when m is constant
so
Force * time = change of momentum, called impulse
impulse = force * time = change of momentum
final momentum = .278 (22.4)
initial momentum = .278 (-16.2)
change = .278 (22.4 - {- 16.2} )
= .278 (38.6)
= 10.7 kg m/s (part a)
force * time = 10.7
force = 10.7/.06 = 179 Newtons (part b)