Standing on the ice, a boy weighing 45 kg throws a stone at an angle of 45 degrees to the horizon at a speed of 5 m / s.

Standing on the ice, a boy weighing 45 kg throws a stone at an angle of 45 degrees to the horizon at a speed of 5 m / s. How fast will the boy drive away if the mass of the stone is 600 grams.

mm = 45 kg.

mk = 600 g = 0.6 kg.

∠α = 45 °.

Vk “= 5 m / s.

Vm “-?

If we neglect the friction between the boy and the ice, then they can be considered a closed system of interacting bodies. For a boy and a stone, the law of conservation of momentum is valid in vector form: mm * Vm + mk * Vk = mm * Vm “+ mk * Vk”.

Since before the throw the boy with the stone was at rest Vm = Vk = 0 m / s, the law will take the form: mm * Vm “+ mk * Vk” = 0.

For projections on the horizontal axis, it will take the form: mm * Vm “+ mk * Vk” * cosα = 0.

Vm “= – mk * Vk” * cosα / mm.

Vm “= – 0.6 kg * 5 m / s * cos45 ° / 45 kg = – 0.046 m / s.

The “-” sign indicates that the boy will move to the opposite side of the stone throw.

Answer: the boy’s speed will be Vm “= 0.046 m / s.