Question

A 47.4 kg student runs down the sidewalk and jumps with a horizontal speed of 42 m/s onto a stationary skateboard
Both the student and skateboard move down the sidewalk with a speed of 3.95 m/s. What's the skateboard's mass?

Answer 1

Answer:

3 kg

Explanation:

Momentum before = momentum after

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

After jumping on the skateboard, the student and the skateboard have the same velocity, so v₁ = v₂.

m₁u₁ + m₂u₂ = (m₁ + m₂) v

(47.4 kg) (4.2 m/s) + m (0 m/s) = (47.4 kg + m) (3.95 m/s)

m = 3 kg

Answer 2

The student and skateboard move down the sidewalk with same speed. By the conservation of momentum, mass of the skateboard is 3 kg,

What is conservation of momentum?

Momentum of a object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity.

When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the law of conservation of momentum.

Thus by the law of conservation of momentum,

$m_1\times u_1+m_2\times u_2=(m_1+m_2)v$

Here, $m_1, m_2$ are the masses of two bodies and $u_1, u_2$ are the initial velocities.

Given information-

The weight of the student is 47.4 kg.

The speed of the student is 42 m/s.

The student and skateboard move down the sidewalk with a speed of 3.95 m/s.

Let the mass of the skateboard is $m_s$ kg.

As the the student and skateboard move down the sidewalk with same speed. Thus by the conservation of momentum,

$47.4\times4.2+m_s\times(0)=(47.4+m_s)(3.95)\\m_s=3\rm kg$

Thus the mass of the skateboard is 3 kg,

Learn more about the conservation of momentum here;

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