A 1400 kg car is traveling east on the highway at 31 m/s and collides into the rear of a slower moving pickup truck of 2400 kg, which is traveling east at 25 m/s. Immediately after the collision, the pickup truck is now traveling at 34 m/s to the east. Determine the speed of the car immediately after the collision.

31 m/s due east

Explanation: this question can be solved using the law of conservation of linear momentum.

This law states that in a closed or isolated system, during collision, the vector sum of momentum before collision equals the vector sum of momentum after collision.

Momentum = mass × velocity

From our question, our parameters before collision are given below as

Mass of car = mc = 1400kg

Speed of car =vc = 31 m/s (due east)

Mass of truck = mt = 2400kg

Velocity of truck = vt = 25 m/s ( due east )

After collision

Velocity of car = ?

Velocity of truck = 34 m/s ( due east )

Vector sum of momentum before collision is given as

1400 (31) + 2400 (25) = 43400 + 60000 = 103400 kgm/s

After collision the truck is seen to move faster (v = 34 m/s) which implies that the car also moves due east .

1400 (v) + 2400(25) A positive value is between both momenta because they are in the same direction.

After collision, we have that

1400v + 60000

Vector sum of momentum before collision = vector sum of momentum after collision

103400 = 1400v + 60000

103400 - 60000 = 1400v

43400 = 1400v

v = 43400/ 1400

v = 31 m/s due east

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