Answer:
The impulse exerted by one cart on the other has a magnitude of 4 N.s.
Explanation:
Given;
mass of the first cart, m₁ = 2 kg
initial speed of the first car, u₁ = 3 m/s
mass of the second cart, m₂ = 4 kg
initial speed of the second cart, u₂ = 0
Let the final speed of both carts = v, since they stick together after collision.
Apply the principle of conservation of momentum to determine v
m₁u₁ + m₂u₂ = v(m₁ + m₂)
2 x 3 + 0 = v(2 + 4)
6 = 6v
v = 1 m/s
Impulse is given by;
I = ft = mΔv = m(
The impulse exerted by the first cart on the second cart is given;
I = 2 (3 -1 )
I = 4 N.s
The impulse exerted by the second cart on the first cart is given;
I = 4(0-1)
I = - 4 N.s (equal in magnitude but opposite in direction to the impulse exerted by the first).
Therefore, the impulse exerted by one cart on the other has a magnitude of 4 N.s.
