bumper car A (281 kg) moving + 2.82 m/s makes an elastic collision with bumper car B (209 kg) moving + 1.72 m/s what is the velocity of car A after the collision (unit = m/s) remember: right is (+), left is (-)

Answer: V1 = 3.559 - 0.744V2

Explanation: Given that the

bumper car A has

Mass M1 = (281 kg) moving with

Velocity U1 = 2.82 m/s

bumper car B has

Mass M2 = (209 kg) moving with

Velocity U2 = 1.72 m/s

Where U1, U2 are the initial velocity of the two cars

Since the collision is elastic, we will use the formula below,

M1U1 + M2U2 = M1V1 + M2V2

Substitute the values into the formula

281*2.28 + 209*1.72 = 281V1 + 209V2

640.68 + 359.48 = 281V1 + 209V2

1000.16 = 281V1 + 209V2

Make V1 the subject of formula

281V1 = 1000.16 - 209V2

V1 = 1000.16/281 - 209V2/281

V1 = 3.559 - 0.744V2

Therefore, the velocity of car A which

is V1 after the collision will be expressed as V1 = 3.559 - 0.744V2