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Momentum problems

I am currently doing my physics homework for momentum and I'm not 100% sure if my answers are right, and I hope someone can confirm it is either correct or not.

Q1) In a simulated accident, a car of mass 800 kg is travelling towards a wall. Just before impact, the speed of the car is 32ms^-1. It rebounds at two-thirds of its initial speed. The car takes 0.50s to come to rest.

(i) Calculate the change in momentum of the car
Δp = m(v-u) Δp = 800((-64/3) - 32) Δp = -42666.67kg ms^-1

(ii) The magnitude and direction of the average force acting on the car during impact
F=(mv-mu)/t F=(-25600)/0.5 F=51200N opposite to initial direction of motion

Q2) A cricketer throws a ball of mass 0.16kg. The ball starts from rest and is thrown horizontally. (The area under the F*t graph is 3.84Ns)
(i) Calculate the speed of the ball, mass 0.16kg when it is released.
v=Δp/m 3.84/0.16 = 24

(ii) Calculate the maximum horizontal acceleration of the ball
I have no idea, I have tried various equations and either gotten a=75 or 150, which seems absurd
Original post by Guarddyyy
I am currently doing my physics homework for momentum and I'm not 100% sure if my answers are right, and I hope someone can confirm it is either correct or not.

Q1) In a simulated accident, a car of mass 800 kg is travelling towards a wall. Just before impact, the speed of the car is 32ms^-1. It rebounds at two-thirds of its initial speed. The car takes 0.50s to come to rest.

(i) Calculate the change in momentum of the car
Δp = m(v-u) Δp = 800((-64/3) - 32) Δp = -42666.67kg ms^-1

(ii) The magnitude and direction of the average force acting on the car during impact
F=(mv-mu)/t F=(-25600)/0.5 F=51200N opposite to initial direction of motion
...


I agree your part (i) and (ii) answer and working.
Reply 2
Original post by Guarddyyy
I am currently doing my physics homework for momentum and I'm not 100% sure if my answers are right, and I hope someone can confirm it is either correct or not.

Q1) In a simulated accident, a car of mass 800 kg is travelling towards a wall. Just before impact, the speed of the car is 32ms^-1. It rebounds at two-thirds of its initial speed. The car takes 0.50s to come to rest.

(i) Calculate the change in momentum of the car
Δp = m(v-u) Δp = 800((-64/3) - 32) Δp = -42666.67kg ms^-1

(ii) The magnitude and direction of the average force acting on the car during impact
F=(mv-mu)/t F=(-25600)/0.5 F=51200N opposite to initial direction of motion

Q2) A cricketer throws a ball of mass 0.16kg. The ball starts from rest and is thrown horizontally. (The area under the F*t graph is 3.84Ns)
(i) Calculate the speed of the ball, mass 0.16kg when it is released.
v=Δp/m 3.84/0.16 = 24

(ii) Calculate the maximum horizontal acceleration of the ball
I have no idea, I have tried various equations and either gotten a=75 or 150, which seems absurd


For the Q2ii) is there some graph or something which shows the force against time?
Original post by Guarddyyy

Q2) A cricketer throws a ball of mass 0.16kg. The ball starts from rest and is thrown horizontally. (The area under the F*t graph is 3.84Ns)
(i) Calculate the speed of the ball, mass 0.16kg when it is released.
v=Δp/m 3.84/0.16 = 24

(ii) Calculate the maximum horizontal acceleration of the ball
I have no idea, I have tried various equations and either gotten a=75 or 150, which seems absurd


As for (ii), it seems that you need info from the graph of F vs time, to find the max force and use F = ma to calculate max horizontal acceleration.
Reply 4
Original post by mqb2766
For the Q2ii) is there some graph or something which shows the force against time?


Original post by Eimmanuel
As for (ii), it seems that you need info from the graph of F vs time, to find the max force and use F = ma to calculate max horizontal acceleration.


This is the graph (link below). I couldn't directly attach it and I'm not 100% sure if I can use imgur links, but I'm sure someone will take it down if it's not allowed. The maximum force was 24N but when I substitute the values into F=ma, 24=0.16a, the acceleration is 150ms^-2 which seems too ridiculous.
https://imgur.com/a/IvPhTmm
(edited 5 years ago)
Original post by Guarddyyy
There was a graph but for some reason I cannot attach it. The maximum force was 24N but when I substitute the values into F=ma, 24=0.16a, the acceleration is 150ms^-2 which seems too ridiculous.


I think it is correct based on what you have described.
Reply 6
Original post by Guarddyyy
This is the graph (link below). I couldn't directly attach it and I'm not 100% sure if I can use imgur links, but I'm sure someone will take it down if it's not allowed. The maximum force was 24N but when I substitute the values into F=ma, 24=0.16a, the acceleration is 150ms^-2 which seems too ridiculous.
https://imgur.com/a/IvPhTmm


Sounds ok to me. If you took an average force of 10, this would be an average acceleration of ~60, and after 0.3 sec, you'd have a final speed of ~20 which fits your other working. So peak acc of 150 seems ok.
Reply 7
Alright, thank you very much everyone for your help! :biggrin:

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