A bag weighing 100 kg is thrown from above into an open car moving by inertia at a speed of 36 km / h

A bag weighing 100 kg is thrown from above into an open car moving by inertia at a speed of 36 km / h, how will the speed of the car change and what will be equal to? Vehicle weight 1000 kg.

Let in an open car weighing m₁ = 1000 kg, moving by inertia with a horizontal speed v₀₁ = 36 km / h = 10 m / s, throw from above a bag weighing m₂ = 100 kg, having a horizontal speed v₀₂ = 0 m / s.

Since the impact of the load with the car was inelastic, that is, v₁ = v₂ = v, then according to the law of conservation of momentum in projection onto the horizontal axis, we obtain:

m₁ ∙ v₀₁ + m₂ ∙ v₀₂ = m₁ ∙ v₁ + m₂ ∙ v₂ or m₁ ∙ v₀₁ = (m₁ + m₂) ∙ v.

Then the vehicle speed becomes: v = m₁ ∙ v₀₁: (m₁ + m₂).

Substitute the values ​​of physical quantities in the calculation formula:

v = 1000 kg ∙ 10 m / s: (1000 kg + 100 kg);

v = 100/11 m / s ≈ 9.1 m / s.

Speed ​​changed to:

v₀₁ – v = 10 m / s – 100/11 m / s = 0.9 m / s.

Answer: the vehicle speed has become ≈ 9.1 m / s; it decreased by 0.9 m / s.