A boy of mass 60 kg jumps with a horizontal velocity of 7.5 ms⁻¹ onto a stationary cart with frictionless wheels. The mass of the cart is 4.5 kg. What is his velocity as the cart starts moving if there is no external unbalanced force working in the horizontal direction?

According to this principle of conservation of momentum, the total momentum of a system remains constant if there is no external unbalanced force acting on it.

Before the jump, the boy is moving with a horizontal velocity of 7.5 m/s, and the cart is stationary. Therefore, the initial momentum of the system is:

p₁ = m₁v₁ + m₂v₂ = (60 kg)(7.5 m/s) + (4.5 kg)(0 m/s) = 450 kg m/s

After the jump, the boy and the cart move together with a common velocity v. Therefore, the final momentum of the system is:

p₂ = (m₁ + m₂) v = (60 kg + 4.5 kg) v = 64.5 kg v

Since there is no external unbalanced force acting on the system in the horizontal direction, the total momentum of the system is conserved:

p1 = p2

Therefore, we have:

450 kg m/s = 64.5 kg v

Solving for v, we get:

v = 6.98 m/s (approximately)

Therefore, the velocity of the boy and the cart after the jump is approximately 6.98 m/s in the horizontal direction.