A skater weighing 68 kg, standing on skates on the ice, throws a 4 kg stone at a speed of 5 m / s at an angle of 30 °

A skater weighing 68 kg, standing on skates on the ice, throws a 4 kg stone at a speed of 5 m / s at an angle of 30 ° to the horizon. What speed does the skater gain immediately after the throw?

m = 68 kg.

mk = 4 kg.

Vk “= 5 m / s.

∠α = 30 °.

V “-?

Since the skater interacts only with the stone, they can be considered a closed system of interacting bodies for which the law of conservation of momentum is valid.

m * V + mk * Vk = m * V “+ mk * Vk” – the law of conservation of momentum in vector form.

Since the skater and the stone were at rest before the throw, their speed before the throw is 0: V = Vk = 0 m / s.

m * V “+ mk * Vk” = 0.

For projections on the horizontal axis, the law of conservation of momentum will take the form: m * V “+ mk * Vk” * cosα = 0.

V “= – mk * Vk” / m * cosα.

The sign “-” indicates that the speed of the skater is directed in the opposite direction of throwing the stone.

V “= 4 kg * 5 m / s / 68 kg * cos30 ° = 0.34 m / s.

Answer: after the throw, the skater will move at a speed V “= 0.34 m / s in the opposite direction of throwing the stone.