Question

When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0 kilogram cart initially at rest, the carts locked together. What is the speed of the combined carts after the collision

Answer 1

The speed of the combined carts after the collision, calculated using the conservation of momentum, is 2.0 meters per second.

Step-by-step explanation:

When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0-kilogram cart initially at rest, and the carts lock together, we can determine the speed of the combined carts after the collision by using the principle of conservation of momentum. Momentum before the collision is equal to the total momentum after the collision since there are no external forces.

The initial momentum of the system can be calculated as the product of the mass and velocity of the moving cart, which gives us (5.0 kg) × (2.8 m/s) = 14.0 kg·m/s (since the second cart is at rest, it contributes no momentum). After the collision, the total mass is the sum of both cart masses, 5.0 kg + 2.0 kg = 7.0 kg. Since momentum is conserved, the final velocity (v) of the combined carts can be found using the equation:

Initial momentum = Final momentum
14.0 kg·m/s = 7.0 kg × v
v = 14.0 kg·m/s ÷ 7.0 kg
v = 2.0 m/s

The speed of the combined carts after the collision is 2.0 meters per second.

Answer 2

Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

`m_1v_(1i) + m_2v_(2i) = m_1v_(1f) + m_2v_(2f)`

as we know that

`m_1 = 5 kg`

`m_2 = 2 kg`

`v_(1i) = 2.8 m/s`

`v_(2i) = 0 m/s`

now from above equation we have

`5(2.8) + 2(0) = m_1v+ m_2v`

`14 = (5+ 2) v`

`v = 2 m/s`

so the speed of combined system is 2 m/s