Question 1
A 0.50 kg football is thrown with a velocity of 15 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.020s . What is the force exerted on the receiver?
So I used mvf-mvi/change(t)
I got 375 for that one...
Question 2
A 0.40 kg soccer ball approaches a player horizontally with a velocity of 18 m/s to the north. The player strikes the ball and causes it to move in the opposite direction with a velocity of 22 m/s. What impulse was delivered to the ball by the player?
I am not sure on this one but I used the formula of: mass * change in velocity and got 1.6 N
Now the last I do not understand and if someone could help me I would really appreciate it:
Question 3
An 82 kg man drops from rest on a diving board 3.0 m above the surface of the water and comes to rest 0.55 after reaching the water. What force does the water exert on him?
I think I may be overthinking this (if that's a word hah) but does gravity play a part here with the 9.8 m/s 2 ?
Please help this assignment is very important to me.
Question 1 is ok but you need to include the force units (Newtons)
In Question 2 you are asked for impulse, not the force. The impulse is 16 Newton-seconds. You missed a decimal point somewhere.
Yes, gravity plays a role in Question 3
Calculte his velocity when he hits the water; sqrt(2gH)= 7.67 m/s
Water exerts a force that (after subtracting the weight)cancels his momentum.
(F - Mg)*0.55 s = M*7.67 m/s)
F = M [g + (7.67 m/s)/0.55 s]
Question 1:
To calculate the force exerted on the receiver, you correctly applied the formula for average force: F = (mvf - mvi) / Δt, where F is the force, m is the mass of the object (in this case, the football), vf is the final velocity, vi is the initial velocity, and Δt is the change in time.
Given:
m = 0.50 kg (mass of the football)
vf = 0 m/s (final velocity, as the ball comes to rest)
vi = 15 m/s (initial velocity)
Δt = 0.020 s (time interval)
Plugging these values into the formula, we have:
F = (0 - 0.50 kg * 15 m/s) / 0.020 s
F = (-7.5 kg·m/s) / 0.020 s
F = -375 N
It seems you made a sign error in your calculation. The negative sign indicates that the force exerted on the receiver is in the opposite direction to the motion of the football. So, the correct force exerted on the receiver is 375 N, not -375 N.
Question 2:
To calculate the impulse delivered to the soccer ball, you correctly used the formula: impulse = mass * change in velocity, where impulse is the change in momentum, mass is the mass of the object (the soccer ball), and the change in velocity is the final velocity subtracted from the initial velocity.
Given:
m = 0.40 kg (mass of the soccer ball)
vf = -22 m/s (final velocity, since the ball moves in the opposite direction)
vi = 18 m/s (initial velocity)
Plugging these values into the formula, we have:
impulse = 0.40 kg * (-22 m/s - 18 m/s)
impulse = 0.40 kg * (-40 m/s)
impulse = -16 N·s
Again, you made a sign error in your calculation. The negative sign indicates that the impulse is in the opposite direction to the initial velocity. However, impulse is a vector quantity, so we consider its magnitude. Therefore, the correct impulse delivered to the ball by the player is 16 N·s, not -16 N·s.
Question 3:
In this scenario, gravity does play a role. The force the water exerts on the man is the net force acting on him, which takes into account the force of gravity and the force exerted by the water.
To find the force exerted by the water, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Given:
m = 82 kg (mass of the man)
g = 9.8 m/s² (acceleration due to gravity)
Δt = 0.55 s (time taken to come to rest)
First, we need to find the initial velocity of the man just before he hits the water. We can use the equation for final velocity in terms of initial velocity, acceleration, and time:
vf = vi + at
Since the man drops from rest, vi (initial velocity) is 0, and the equation simplifies to:
vf = at
Substituting the known values, we have:
0 m/s = (9.8 m/s²) * 0.55 s
0 m/s = 5.39 m/s
Now, we can calculate the deceleration of the man when hitting the water. We can use the equation for average acceleration:
Δv = vf - vi = -5.39 m/s - 0 m/s = -5.39 m/s
a = Δv / Δt = (-5.39 m/s) / 0.55 s
a ≈ -9.798 m/s²
Since the man comes to rest, his acceleration is in the opposite direction to his initial velocity, and that is why it is negative.
Finally, we can calculate the force exerted by the water using Newton's second law:
F = m * a
F = 82 kg * (-9.798 m/s²)
F ≈ -803.636 N
Again, the negative sign indicates that the force exerted by the water is in the opposite direction to the initial motion of the man. Hence, the correct force exerted by the water on the man is approximately 803.636 N, not -803.636 N.
For Question 1:
To calculate the force exerted on the receiver when catching the ball, you can use Newton's second law: F = Δp/t, where F is the force, Δp is the change in momentum, and t is the time taken to bring the ball to rest.
Given:
Mass of the football (m) = 0.50 kg
Initial velocity of the football (vi) = 15 m/s (to the right)
Final velocity of the football (vf) = 0 m/s
Time taken to bring the ball to rest (t) = 0.020 s
The change in momentum (Δp) can be calculated using: Δp = m(vf - vi).
Δp = 0.50 kg (0 m/s - 15 m/s)
Δp = -7.5 kg·m/s (since the momentum is in the opposite direction)
Now, we can calculate the force exerted on the receiver:
F = Δp / t
F = (-7.5 kg·m/s) / 0.020 s
F ≈ -375 N (since the force is in the opposite direction)
Therefore, the force exerted on the receiver is approximately -375 N (to the left).
For Question 2:
To calculate the impulse delivered to the soccer ball by the player, you can use the formula: Δp = m(vf - vi), where Δp is the change in momentum, m is the mass of the soccer ball, and (vf - vi) is the change in velocity.
Given:
Mass of the soccer ball (m) = 0.40 kg
Initial velocity of the soccer ball (vi) = 18 m/s (to the north)
Final velocity of the soccer ball (vf) = -22 m/s (to the south)
The change in momentum (Δp) can be calculated using: Δp = m(vf - vi).
Δp = 0.40 kg (-22 m/s - 18 m/s)
Δp = -16 kg·m/s (since the momentum is in the opposite direction)
Therefore, the impulse delivered to the soccer ball by the player is -16 kg·m/s.
Now, for Question 3:
When the man drops from rest, there are two forces acting on him: the force of gravity pulling him down and the force exerted by the water resistance. In this case, we need to calculate the force exerted by the water on the man.
Using the equation of motion: vf^2 = vi^2 + 2ad, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the distance.
To find the acceleration (a), we can rearrange the equation as: a = (vf^2 - vi^2) / (2d).
Given:
Mass of the man (m) = 82 kg
Initial velocity of the man (vi) = 0 m/s
Final velocity of the man (vf) = 0 m/s
Distance fallen (d) = 3.0 m
Time taken to come to rest (t) = 0.55 s
The acceleration (a) can be calculated using: a = (vf^2 - vi^2) / (2d).
a = (0 m/s - 0 m/s) / (2 × 3.0 m)
a = 0 m/s^2
Since the final velocity is 0 m/s, it means that the man comes to rest completely. Hence, there is no acceleration.
Now, we can calculate the force exerted by the water on the man:
Using Newton's second law: F = ma, where F is the force, m is the mass, and a is the acceleration.
F = m × a
F = 82 kg × 0 m/s^2
F = 0 N
Therefore, the force exerted by the water on the man is 0 N.