Question

Cart A (50. kg) approaches cart B (100. kg initially at rest) with an initial velocity of 30. m/s. After the collision,
cart A locks together with cart B. both travels with what
velocity?​

Answer 1

Given:-

→Mass of Cart A(m₁) = 50kg

→Mass of Cart B(m₂) = 100kg

→Initial velocity of Cart A(u₁) = 30m/s

→Initial velocity of Cart B(u₂) = 0

→After collision , Cart A locks Cart B

To find:-

→Velocity of the combined object after

collision.(v)

Solution:-

Here in this case, we shall apply Law Of Conservation of momentum, to get the required answer:-

=> m₁u₁ + m₂u₂ = (m₁+m₂)v

=> 50(30)+100(0) = (50+100)v

=> 1500 = 150v

=> v = 1500/150

=> v = 10m/s

Thus, after collision both will travel with a velocity of 10m/s.

Some Extra Information:-

We have studied that.

=> F = m(v-u)/t

=> Fₑₓₜ = mv-mu/t [∵F = Fₑₓₜ]

When Fₑₓₜ = 0

=>0×t = mv-mu

=>mv-mu = 0

=>mv = mu

The final momentum is equal to the initial momentum i.e. momentum remains conserved.

Thus Conservation of momentum can be stated as "when external force acting on a system is zero, then total momentum of the system remains constant".

Answer 2

Given :-

☄ Mass of Cart A, m₁ = 50kg

☄ Mass of Cart B, m₂ = 100kg

☄ Initial velocity of Cart A, u₁ = 30m/s

☄ Initial velocity of Cart B, u₂ = 0

To find:-

Velocity of the carts after collision.

Solution:-

we know that cart A locks Cart B after collision.

using Law Of Conservation of momentum .i.e.,

➠ m₁u₁ + m₂u₂ = (m₁+m₂)v

➠ 50(30)+100(0) = (50+100)v

➠ 1500 + 0 = 150v

➠ v = 1500/150

➠ v = 10m/s

Thus, after collision both carts travel with a velocity of 10m/s.

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