# Momentum, Impulse and collisions Q

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:

CAF123
Gold Member
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?

Hi CAF123,

ehild
Homework Helper
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

Hi ehild,

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

ehild
Homework Helper
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 the 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

1. ΔP=Pfinal-Pinital
∴(2200*10)+(1200*15)=40,000 Newtons
Since P=mv ∴v=P/m ∴ 40,000 / (1200+2200) = 11.76 m/s
2. Q2
Since F=m*a ∴ F=2200*a
where a = (v-u)/t ∴ (11.76-10)/0.4 = 4.41 m/s^2

thus F=2200*4.41 = 9705.88 Newtons
3. Same as q2, 9705.88 Newtons

Proof:
F=1200*a
where a = (v-u)/t ∴ (11.76-15)/0.4 = -8.088 m/s^2

thus F=2200*-8.088 = -9705.88 Newtons

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

ehild
Homework Helper
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?

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

ehild

CAF123
Gold Member
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.

(By the way, the unit of momentum is Ns or kgm/s, not Newton)
Thanks for the clarification. Does "Ns" actually stand for anything?

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

ehild
Homework Helper
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

CAF123
Gold Member
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.

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

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

CAF123
Gold Member
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 $v_2$

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

CAF123
Gold Member
PS: How come you guys didn't work it out step-by-step for me? ;o
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.

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 . .

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.

yup calculation seems alright

CAF123
Gold Member
1.Total P= m1v1 + -(m2v2) car in opposite direction
∴(2200*10)-(1200*15)=4,000 Newtons

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