a) A tennis racket a 450 N force on a 0.05 kg tennis ball over a 0.006 s time period. What is the impulse applied to the ball?
b) A ball (0.5 kg) is thrown with a velocity of 15 m/s. A receiver catches the ball and brings it to rest is 0.02 s.
a) What is the change in momentum?
c) A ball (0.5 kg) is thrown with a velocity of 15 m/s. A receiver catches the ball and brings it to rest is 0.02 s.
a) What is the change in momentum of the ball?
b) What is the impulse delivered to the ball?
c) What force is exerted on the ball?
d) A ball (0.40 kg) is moving with a velocity of 18 m/s. A player strikes the ball and causes it to move in the opposite direction with a velocity of 22 m/s. The player exerts a force of 5500 N.
a) What is the change in momentum of the ball?
b) What is the impulse delivered to the ball?
c)How long is the foot in contact with a ball?
~formula's~
J= Ft
J= P
Ft=mvf-mvi
PLease help me!! :)
These are all the same so I am only going to do d
initial momentum = .4 * 18 = + 7.2 kg m/s
final momentum = - .4 * 22 = - 8.8 kg m/s
change in momentum = -8.8 - 7.2 = -16 kg m/s
impulse = change in momentum = -16 kg m/s
Force = change in momentum / change in time
so
-5500 = -16 /t
t = .003 seconds
a) To calculate the impulse applied to the ball, we can use the formula J = Ft, where J is the impulse, F is the force, and t is the time period.
Given:
Force on the ball (F) = 450 N
Time period (t) = 0.006 s
Using the formula J = Ft, we can calculate the impulse:
J = 450 N * 0.006 s
J = 2.7 Ns
So, the impulse applied to the ball is 2.7 Ns.
b) To calculate the change in momentum, we can use the formula Δp = mvf - mvi, where Δp is the change in momentum, m is the mass, vf is the final velocity, and vi is the initial velocity.
Given:
Mass of the ball (m) = 0.5 kg
Initial velocity (vi) = 15 m/s
Final velocity (vf) = 0 m/s (brought to rest)
Using the formula Δp = mvf - mvi, we can calculate the change in momentum:
Δp = 0.5 kg * 0 m/s - 0.5 kg * 15 m/s
Δp = -7.5 kgm/s
So, the change in momentum is -7.5 kgm/s. Note that the negative sign indicates a reversal in direction.
c) Continuing from the previous question, now we have the following:
Change in momentum (Δp) = -7.5 kgm/s
To calculate the impulse delivered to the ball, we can use the formula J = Δp.
Using the given value of Δp, we have:
J = -7.5 kgm/s
So, the impulse delivered to the ball is -7.5 kgm/s. Again, note that the negative sign indicates a reversal in direction.
d) To calculate the force exerted on the ball, we can rearrange the formula J = Ft to solve for F.
Given:
Impulse delivered to the ball (J) = -7.5 kgm/s
Time period (t) = 0.02 s
Using the rearranged formula F = J/t, we can calculate the force:
F = -7.5 kgm/s / 0.02 s
F = -375 N
So, the force exerted on the ball is -375 N. Note that the negative sign indicates a reversal in direction.
e) Given:
Mass of the ball (m) = 0.40 kg
Initial velocity (vi) = 18 m/s
Final velocity (vf) = -22 m/s (moving in the opposite direction)
Force exerted on the ball (F) = 5500 N
a) To calculate the change in momentum, we can use the formula Δp = mvf - mvi, where Δp is the change in momentum, m is the mass, vf is the final velocity, and vi is the initial velocity.
Δp = 0.40 kg * (-22 m/s) - 0.40 kg * 18 m/s
Δp = -8.8 kgm/s - 7.2 kgm/s
Δp = -16 kgm/s
So, the change in momentum of the ball is -16 kgm/s.
b) To calculate the impulse delivered to the ball, we can use the formula J = Δp.
J = -16 kgm/s
So, the impulse delivered to the ball is -16 kgm/s.
c) To calculate the duration of contact, we can use the formula J = Ft and rearrange it to solve for t.
J = Ft
t = J / F
t = -16 kgm/s / 5500 N
t ≈ -0.0029 s
Therefore, the duration of contact is approximately 0.0029 seconds. Note that the negative value indicates a reversal in direction.
Sure! I'd be happy to help you with these questions using the given formulas.
a) To find the impulse applied to the tennis ball, we can use the formula J = Ft, where J is the impulse, F is the force, and t is the time period. In this case, the force is 450 N and the time period is 0.006 s. So, the impulse applied to the ball is J = (450 N) * (0.006 s) = 2.7 Ns.
b) To find the change in momentum, we can use the formula ΔP = mvf - mvi, where ΔP is the change in momentum, m is the mass, vf is the final velocity, and vi is the initial velocity. In this case, the mass is 0.5 kg, the initial velocity is 15 m/s, and the final velocity is 0 m/s (since the ball is brought to rest). So, the change in momentum is ΔP = (0.5 kg) * (0 m/s - 15 m/s) = -7.5 kg m/s (negative because the momentum is decreasing).
c) To find the change in momentum of the ball, we already calculated it as -7.5 kg m/s in part b.
To find the impulse delivered to the ball, we can use the formula J = ΔP, where J is the impulse and ΔP is the change in momentum. In this case, the impulse delivered to the ball is J = -7.5 kg m/s.
To find the force exerted on the ball, we can rearrange the formula J = Ft to solve for force: F = J / t, where F is the force and t is the time period. In this case, the impulse is -7.5 kg m/s and the time period is 0.02 s. So, the force exerted on the ball is F = (-7.5 kg m/s) / (0.02 s) = -375 N (negative because the force is acting in the opposite direction).
d) To find the change in momentum of the ball, we can use the same formula as in part b: ΔP = mvf - mvi. In this case, the mass is 0.40 kg, the initial velocity is 18 m/s, and the final velocity is -22 m/s (since the ball is moving in the opposite direction). So, the change in momentum is ΔP = (0.40 kg) * (-22 m/s - 18 m/s) = -16 kg m/s (negative because the momentum is decreasing).
To find the impulse delivered to the ball, we can use the formula J = ΔP. In this case, the impulse delivered to the ball is J = -16 kg m/s.
Unfortunately, the information needed to answer the question about the duration of contact between the ball and the player's foot is not provided. However, if you are given the average force exerted during the collision, you can use the formula F = ΔP / t, where F is the force, ΔP is the change in momentum, and t is the time of contact, to find the duration of contact (t).