How Do You Solve a Momentum Problem Involving Friction and Collision?

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The discussion addresses a momentum problem involving a collision between a small car (mass = 500 kg) and a sports car (mass = 1000 kg) after the small car skids for 40 meters due to a constant frictional force of 7.5 x 108 Newtons. Key calculations include determining the deceleration of the two cars using the formula F = ma, leading to a deceleration of μg. The conservation of momentum principle is applied to find the speed of the sports car just before impact and the speed of both cars post-collision.

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Here's the problem:
A small car (mass = 500 kg) fails to stop at the entrance to a major road until it has
reached the middle of the carriageway. It is then struck by a sports car (mass =
1000 kg) and the two cars lock together. Police investigators show that the skid
marks stretch for 40 m. Assuming that the frictional force on these two vehicles
was constant at 7.5 x 108 Newtons, find

(a) the deceleration of the two cars
(b) the speed at which the two cars moved off after the impact
(c) the speed of the sports car just before the impact…in miles per hour (100 mi/h = 45
m/s approx.)

I couldn't apply any formula of momentum here.there's even no given time and velocity.i tried to apply F=ma to get the deceleration but i think constant friction is the key here.a also have no idea how to get the other two questions.i'm kinda confused here.help please..i'd appreciate a good explanation how to solve this problem.thanks a lot.
 
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As you said, F=ma and the constant friction force are the key to part A. So, you know:

The friction force and \Sigma F=ma.

Can you use this to find anything of interest?
 
are you sure the formula is applicable?
i think the answer would be -500000 m/s2
isn't that too unreal?
how bout the constant friction?is it related to the acceleration?
thanks.
 
Momentum is conserved

p_{1v1} = momentum of vehicle 1 prior to collision
p_{1v2} = momentum of vehicle 2 prior to collision
p_{2v1} = momentum of vehicle 1 after to collision
p_{2v2} = momentum of vehicle 2 after to collision

p_{1v1}+p_{1v2}=p_{2v1}+p_{2v2}
 
You also know how to convert between distance and acceleration:

a(x-x_0)=\frac{1}{2}(v^2-v_0^2)

Using this you can find the initial velocity v_0

You know the friction force is μN=μmg and since m cancels out, the deceleration is μg
 
Last edited:

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