Question

A 60-kilogram man holding a 20-kilogram box rides on a skateboard at a speed of positive 7 meters per second. He throws the box behind him, giving it a velocity of negative 5 meters per second, with respect to the ground. What is his velocity, in meters per second, after throwing the object?

Answer 1

Answer:

$v_1=11\ m/s$

Explanation:

It is given that,

Mass of the man, m₁ = 60 kg

Mass of the box, m₂ = 20 kg

Speed of man+ box, V = +7 m/s

Speed of the box, when it was thrown, v₂ = -5 m/s

To find,

Velocity of the man after throwing the box

Solution,

Let v₁ be the velocity of the man after throwing the box. By using the conservation of momentum, it can be calculated.

$(m_1+m_2)V=m_1v_1+m_2v_2$

$(60+20)\times 7=60\times v_1+20\times (-5)$

$v_1=11\ m/s$

Therefore, the velocity of the man after throwing the object is 11 m/s towards his initial direction.