Given
m1(mass of red bumper): 225 Kg
m2 (mass of blue bumper): 180 Kg
m3(mass of green bumper):150 Kg
v1 (velocity of red bumper): 3.0 m/s
v2 (final velocity of the combined bumpers): ?
The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:
Pa= Pb
Where Pa is the momentum before collision and Pb is the momentum after collision.
Now applying this law for the above problem we get
Momentum before collision= momentum after collision.
Momentum before collision = (m1+m2) x v1 =(225+180)x 3 = 1215 Kgm/s
Momentum after collision = (m1+m2+m3) x v2 =(225+180+150)x v2
=555v2
Now we know that Momentum before collision= momentum after collision.
Hence we get
1215 = 555 v2
v2 = 2.188 m/s
Hence the velocity of the combined bumper cars is 2.188 m/s
Answer: B. 1.2 m/s
Explanation:
First we would find the velocity with which red car and blue car move when they combine after first collision.
Using law of conservation of momentum,
Initial momentum = final momentum
Mass of red Car, Mr = 225 kg
Mass of blue car, Mb = 180 kg
Mass of green car, Mg = 150 kg
Initial velocity of red car, Ur = 3.0 m/s
Initial velocity of blue car, Ub = 0
Initial velocity of green car, Ug = 0
collision 1:
Mr Ur + Mb Ub = (Mr+Mb) V
where, V is the final velocity with which red and blue car moves.
225 × 3 + 0 = (225 + 180 ) V
⇒ V = 675/ 405 = 1.67 kg.m/s
collision 2:
These cars collide with the green car. Let the final velocity with which the three cars move together after collision: V'
(Mr+Mb) V + Mg Ug = (Mr+Mb+Mg) V'
675 + 0 = 555 V'
⇒V' = 1.2 m/s
Thus, the correct option is B.
