Car Collision Speed Calculations

AI Thread Summary
In a collision involving a 1.2-tonne car and a stationary 2.4-tonne van, the vehicles became entangled and skidded 15 meters before stopping, with a friction force of 2880 N. To find the speed of the vehicles just after the collision, the equations of motion and conservation of momentum were applied, leading to a calculated speed of approximately 4.90 m/s. The initial speed of the car before the collision was determined to be about 14.7 m/s. The calculations involved using the work done by friction and the loss of kinetic energy. The results demonstrate the application of physics principles in analyzing vehicle collisions.
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Homework Statement



A car of mass 1.2 tonnes collides with a stationary van of mass 2.4 tonnes. After the collision the two vehicles become entangled and skid 15 m before stopping. Police accident investigators estimate that the magnitude of the friction force during the skid was 2880 N. Assume the road is horizontal and that all the motion takes place in a straight line.

a) Find the speed of the vehicles just after the collision.

b) Find the speed of the car before the collision.

Homework Equations



m_1u_1 + m_2u_2 = (m_1+m_2)v

p = mv

v^2 = u^2 + 2as

The Attempt at a Solution



a) 1200u1 + 0 = 3600v

to find out the velocity after the collision, use v as the initial velocity to travel 15m with a frictional force of 2880 N.

v2 = u2 + 2as

0 = u^2 + 2as

0 = u^2 + 30a

f = ma

f/m = a

(3600v - 2880)/3600 = a

v - 288/360 = a

u^2 + 30(v-288/360) = 0
 
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you can use conservation of energy for the first part:
The work done by friction to stop the vehicles= loss in kinetic energy of the car-van system
 
LaTeX Code: (3600v - 2880)/3600 = a

LaTeX Code: v - 288/360 = a

This step is wrong. Check it.
 
-2880/3600 = a

0 = u^2 + 30(-2880/3600)
24 = u^2
u = 4.90

b) m_1u_1 + m_2u_2 = (m_1 + m_2)v
1200u_1 = 3600(4.898)
u_1 = 14.7ms^-^1
 
That is right.
 
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