Answer:
Explanation:
consider the mass of each train car be m
m₁ = m₂ = m₃ = m
speed of the three identical train
u₁ = u₂ = u₃ = 1.8 m/s
m₄ = m u₄ = 4.5 m/s
m₅ = m u₅ = 0 (initial velocity )
final velocity
v₁ = v₂ = v₃ = v₄ = v₅ = v
using conservation of momentum
m₁u₁ + m₂u₂ + m₃u₃ + m₄u₄ + m₅u₅ = m₁v₁ + m₂v₂ + m₃v₃ + m₄v₄ + m₅v₅
m (1.8 + 1.8 + 1.8 +4.5) = 5 m v
Answer:
<em>The velocity of the carts after the event is 1 m/s</em>
Explanation:
<u>Law Of Conservation Of Linear Momentum </u>
The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is
P=mv.
If we have a system of bodies, then the total momentum is the sum of the individual momentums:
If a collision occurs and the velocities change to v', the final momentum is:
Since the total momentum is conserved, then:
P = P'
In a system of two masses, the equation simplifies to:
If both masses stick together after the collision at a common speed v', then:
The common velocity after this situation is:
The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:
The velocity of the carts after the event is 1 m/s