# collision, how fast was one car?

• November 14th 2007, 03:53 PM
rcmango
collision, how fast was one car?
A car (mass = 1070 kg) is traveling at 32 m/s

when it collides head-on with a sport utility vehicle (mass = 2390 kg) traveling in the opposite direction.

In the collision, the two vehicles come to a halt.

At what speed was the sport utility vehicle traveling?
in m/s

not sure how to solve this one.
• November 14th 2007, 05:26 PM
mathmonster
work out momentum of each vehicle before the collision!

so say A is the car, A has momentum 32 x 1070

so B is the SUV, and has momentum -v x 2390 (- v because in opposite direction to A's velocity)

so the total momentum is (32 x 1070) - (v x 2390)

now work out momentum after collision

as they come to a hault, then the velocity is zero, and assuming they stay together, the the masses can be added.

so final momentum is 0 x 3460

now put into an equation, because of the conservation of momentum law

we have,

(32 x 1070) - (v x 2390) = 0 x 3460

so after calculation

v = 14.326 m/s
• November 15th 2007, 09:36 PM
rcmango
ah, yes. I see, i put -14.3 for the velocity and that wasn't correct. It must be positive velocity, i must've switch the masses wrong, and subtracted them.

thanks for the help, and solution.
• November 15th 2007, 10:06 PM
Aryth
Let's see how this is done symbolically, that way, we can find out exactly where the negative disappeared:

Well, we have a car (m = 1070kg) and an SUV (m = 2390kg), each going in opposite directions. Well we'll specify -----> as the positive x-direction, so we have:

car ----------------------------> <------------------------------SUV

As you can see, the SUV is going in the negative direction, therefore it's velocity in this will be negative, so that:

$m_1v_1 + m_2v_2 = m_1v_3 + m_2v_4$

Since the two vehicles halt immediately after the collision, we can reduce the entire right side of the equation to zero. And since the SUV's velocity was negative, it's momentum is also negative, therefore the $m_2v_2$ will become $-m_2v_2$. Making the equation:

$m_1v_1 - m_2v_2 = 0$

Now we just simplify by moving the $-m_2v_2$ over, getting:

$m_1v_1 = m_2v_2$

Now we plug in the numbers, remember, we moved the momentum over and it lost its negative, therefore the negative in the velocity is eliminated as well since we pulled it out before plugging the numbers in:

$1070*32 = 2390*v_2$

$34340 = 2390*v_2$

$14.3 \frac{m}{s} = v_2$