Question

A 1500kg car traveling east with speed 25 m/s collides and sticks to another car traveling north. The combined wreck skids to a stop. The skid marks are 13m long and make an angle 23° north of east. Determine the mass and speed of the northbound car. You will need to look up the appropriate coefficient of friction. Cite sources.

Answer 1

(a) As shown in the figure
Initial momentum in the x direction
= m₁V₁ = 1500*25 = 37500 kg-m/s
Initial momentum in the Y direction
= mv --------------(1)
Now after the collision
Momentum along x direction = (1500+m)VCos23 --------(2)
Momentum along Y direction = (1500+m)VSin23 -----------(3)
Now let us assume that the kinetic friction coefficient is u_k = 0.3
Hence
the initial energy of the car after the collision
= (1/2)(1500+m)V²
which will be equal to the friction energy = friction force*distance of skid
friction force= u_K(1500+m)*g
hence
(1/2)(1500+m)V² = u_K(1500+m)*g *13
V = 8.75 m/s
Now applying conservation of momentum along X direction
(1500+m)VCos23 = 37500
(1500+m)*8.75Cos23 = 37500
m = 3155.83 kg
Now in y direction
mv = (1500+m)VSin23
3155.83v = (1500+3155.83)*8.75Sin23
v = 5.044 m/s