# Momentum, Impulse and collisions Q

1. Aug 27, 2012

### leechyeah

A car of mass 1200kg travelling at a velocity of 15m/s North collides head on with a small van of mass 2200kg travelling at 10m/s South. The vehicles stick together on impact and the collisions can be considered an isolated one. The contact time of the collision is 0.4 seconds.
1. Calc the velocity of thethe vehicles after the collision.
2. Calc the average force of the van on the car.
3. State average force of the car on the van.

• F=m*a
• F=(mass*Δvelocity)/Δtime
• momentum=mass*velocity
• ΔP=Pfinal-Pinital (note: p is momentum)
• Total P= m1v1 + m2v2

The following calculations are correct now.
1. Total P= m1v1 + -(m2v2) car in opposite direction
∴(2200*10)-(1200*15)=4,000 Ns(Newton secundum)
Since P=mv ∴v=P/m ∴ 4,000 / (1200+2200) = 1.17 m/s
2. Q2
Since F=m*a ∴ F=2200*a
where a = (v-u)/t ∴ (1.176-10)/0.4 = -22.06 m/s^2

thus F=2200*-22.06 = -48,529.4 Newtons
3. Same as q2, 48529.4 Newtons but in opposite direction, hence the +/- sign.

Proof:
F=1200*a
where a = (v-u)/t ∴ (1.176-(-15))/0.4 = 40.4412 m/s^2

thus F=1200*40.4412 = 48529.4 Newtons

Last edited: Aug 27, 2012
2. Aug 27, 2012

### CAF123

I am not really sure I understand your attempt.
Q1 refers to a velocity as an answer and you have a force.
Have you got an answer for this question?

3. Aug 27, 2012

### leechyeah

Hi CAF123,

Thanks for your help, I've updated my calculations now. I'm not sure about if it's right though. Please help me check.

4. Aug 27, 2012

### ehild

Hi leechyeah, welcome to PF.

Q2 asks the force the van exerts on the car. How long does that force act?
You have written that the change of the momentum of a body is equal to the force exerted on it multiplied by the time during the force acts. ΔP=FΔt.
The momentum of the car has changed. By how much?

ehild

5. Aug 27, 2012

### leechyeah

Hi ehild,

I understand that there is something wrong, but I am confused as to what it is.
Shall I use t=0.4 secs, given in the question?

6. Aug 27, 2012

### ehild

The total momentum of the cars is conserved during collision. The total momentum is P=m1v1+m2v2 but the velocity v and the momentum P are vector quantities. The car and the van travelled in opposite directions, one of the velocities positive, the other is negative. Take that into account.
(By the way, the unit of momentum is Ns or kgm/s, not Newton)

ehild

7. Aug 27, 2012

### ehild

Yes, that you used correctly. The total momentum was wrong.

ehild

8. Aug 27, 2012

### CAF123

Use conservation of momentum to find the combined velocity of the vehicles:
$m_1v_1 + m_2v_2 = (m_1 +m_2)V,$ where $V$ is the velocity you are looking for. Note in your attempt, you forgot that velocity was a vector.

9. Aug 27, 2012

### leechyeah

Thanks for the clarification. Does "Ns" actually stand for anything?

I updated my original post. However, Q3 is where I'm unsure of.

10. Aug 27, 2012

### ehild

Ns is newton-secundum.

As for Q3 think of Newton's third law. If the van exerts F force on the car, how big is the force the car exerts on the van?
Of course, you can also calculate the change of momentum of the van, and divide it by 0.4 s.

ehild

11. Aug 27, 2012

### CAF123

The combined velocity is now correct.
The reason you are getting different forces now is due to your signs. You know they should be equal because of Newton's 3rd. Retry this but using the right signs.

'Ns' is a newton second.

12. Aug 27, 2012

### leechyeah

Thanks,
I understand the net force for q2 and q3 must be the same, however my calculations do not match. Which part is wrong?

13. Aug 27, 2012

### leechyeah

in q1, If I use 11.76 instead of 1.176, then everything else will work out...

14. Aug 27, 2012

### CAF123

Ok, so $F = \frac{m(V-v_1)}{t}.$ $V$ here is always ≈ -1.2m/s. You just need to make sure the signs are right for $v_1$ and $v_2$

15. Aug 27, 2012

### leechyeah

Thanks got it, -15 m/s.
I will amend the original post.

PS: How come you guys didn't work it out step-by-step for me? ;o

16. Aug 27, 2012

### CAF123

If you mean provide a step by step solution, then all I can say is that we are not allowed to do that. Besides, it is a lot more satisfying solving a problem yourself and it is better for you.

17. Aug 27, 2012

### kushan

Its the beauty of PF and specially Mr Ehild ,
They will push you enough so that you come out with answer yourself , and this is exactly the right way of teaching physics . .

18. Aug 27, 2012

### leechyeah

I see,

Thank you all for your help. Lastly, do each of you agree my with my final solution?

I don't have the text book solutions at the moment.

19. Aug 27, 2012

### hav0c

yup calculation seems alright

20. Aug 27, 2012

### CAF123

Your solution is correct. Just to point out that in the above line of working, the units should be kgm/s, not Newtons.