Question

A 1500-kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2200-kg SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.75, and the cars slide to a halt at a point 5.39 m west and 6.43 m south of the impact point. How fast was each car traveling just before the collision?

Answer 1

Answer:

Velocity of Sedan = 21 m/s

Velocity of SUV = 12 m/s

Explanation:

As we know that deceleration due to friction force is given as

$a = - \mu g$

so we have

$a = -(0.75)(9.81)$

$a = -7.36 m/s^2$

now the two cars comes to rest at a point which is at position of 5.39 m West and 6.43 m South

so net displacement of the car is given as

$d = \sqrt{5.39^2 + 6.43^2}$

$d = 8.39 m$

now the velocity of the two cars just after the impact is given as

$v^2 - v_i^2 = 2a d$

$0 - v_i^2 = 2(-7.36)(8.39)$

$v_i = 11.11 m/s$

direction of the motion is given as

$tan\theta = \frac{6.43}{5.39}$

$\theta = 50 degree$ South of West

now we can use momentum conservation as there is no external force on it

Momentum conservation in North to south direction

$m_1 v_1 = (m_1 + m_2) vsin\theta$

$1500 v_1 = (1500 + 2200) (11.11) sin50$

$v_1 = 21 m/s$

Similarly momentum conservation towards West direction

$m_2 v_2 = (m_1 + m_2) vsin\theta$

$2200 v_2 = (1500 + 2200) (11.11) cos50$

$v_2 = 12 m/s$