Question

Two electric model trains are sitting on a long stretch of track. One train has a mass of 5.50 kg and rolls into the second train which is at rest. The second train is much smaller, with a mass of 550.0 g. After the bump, the large and small trains have velocities of 3.2727 cm/s and 7.2727cm/s , respectively. What is the original velocity of the large train before the bump?

Answer 1

The original velocity of the large train before the bump will be 3.999 cm/s.

Velocity can be defined as the rate of change distance with respect to time.

• Its SI unit is m/s.

According to the law of conservation of momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m₁ = mass of the first model electric train = 5.50 kg = 5500 g

u₁ = initial velocity of the first model train

m₂ = mass of the second model electric train = 550.0 g

u₂ = 0 m/s as it is at rest

v₁ = final velocity of the first model train after collision = 3.2727 cm/s

v₂ = final velocity of the second model train after collision = 7.2727 cm/s

By putting the values in the main equation, u₁ can be calculated,

5500*u₁ + 550*0 = 5500*3.2727 + 550*7.2727

5500*u₁ = 17999.85 + 3999.985

5500*u₁ = 21999.835

u₁ = 21999.835/5500

u₁ = 3.999 cm/s

The original velocity of the large train before the bump will be 3.999 cm/s.

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