A thrown ball weighing 2 kg hits a cart weighing 1 kg, the cart starts moving at a speed of 2 m / s

A thrown ball weighing 2 kg hits a cart weighing 1 kg, the cart starts moving at a speed of 2 m / s, and the ball bounces in the opposite direction at a speed of 2 m / s. Determine the speed at which the ball was thrown.

mm = 2 kg.

mt = 1 kg.

Vm “= 2 m / s.

Vt “= 2 m / s.

Vt = 0 m / s.

Vm -?

Let us write the law of conservation of momentum in vector form: mm * Vm + mt * Vt = mm * Vm “+ mt * Vt”, where mm * Vm, mm * Vm “is the impulse of the ball before and after hitting the cart, mt * Vt, mt * Vт “- the impulse of the cart before and after the ball hit.

For projections on the horizontal axis OX, directed towards the ball throwing, the impulse conservation law will take the form: mm * Vm = – mm * Vm “+ mt * Vt”.

Vm = (mt * Vt “- mm * Vm”) / mm.

Vm = (1 kg * 2 m / s – 2 kg * 2 m / s) / 2 kg = – 1 m / s – the speed of the ball cannot be negative.

An error was made in the problem statement.

After the impact, the sum of the impulses of the ball and the trolley is mt * Vt “- mm * Vm” = 1 kg * 2 m / s – 2 kg * 2 m / s = – 2 kg * m / s.

The “-” sign indicates that even before the impact, the total impulse must be directed in the opposite direction of the ball’s flight.

Answer: a mistake was made in the problem statement.