A 750kg car is stalled on an icy road during a snowstorm. A 1000kg car traveling

Question

A 750kg car is stalled on an icy road during a snowstorm. A 1000kg car traveling eastbound at 10 m/s collides with the rear of the stalled car. After being hit, the 750 kg car slides on the ice at 4 m/s in a direction 30 degrees north of east. a) What are the magnitude and direction of the velocity of the 1000 kg car after the collision? b) Calculate the ratio of the kinetic energy of the two cars just after the collision to that just before the collision ( you may ignore the effects of friction during the collision) Please show me the steps. Thank you in advance

Explanation / Answer

here,

m1 = 1000 kg
v1 = 10 m/s

m2 = 750kg
v2'x = 4*cos30 = 4 * 0.866 = 3.464 m/s
V2'y = 4*Sin30 = 4 * 0.5 = 2 m/s

v1'x = v1'CosA
v1'y = V1'SinA

From Conservatin of Momentum we have :

in X direction :
m1v1x = m1v1'x + m2v2'x
1000 * 10 = 1000*v1'x + 3.464 * 750
v1'x = 7.402 m/s

Similarly in y direction :

m1v1y = m1v1'y + m2v2'y

m1v1'y = - m2v2'y
v1'y = - (m2/m1) * v2'y
v1'y = - (750/1000) * 2
V1'y = - 1.5 m/s

Vnet = sqrt(vx^2 + vy^2)
Vnet = sqrt((7.402)^2 + (-1.5)^2)
Vnet = 7.7552 m/s

also, diretion of m1 is given as :

A = arcTan(vy/vx)
A = arcTan (1.5/7.402)
A = 11.46 degrees towards south of east

PART B:
KEf = 0.5m1*v1'^2 + 0.5m2*v2'^2
KEf = 0.5 * 1000 * 7.552^2 + 0.5 * 750 * 4^2
KEf = 34516.352 J

KEi = 0.5 m1*v1^2
KEi = 0.5 * 1000 * 10^2
KEi = 50000 J

Ratio :

KEf/KEi = 34516.352 / 50000 = 0.690

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