Question

A 55 kg boy running at 2.0 m/s jumps onto a 2.0 kg skateboard. Calculate the final velocity of the boy and the skateboard.​

Answer 1

Answer:

The final velocity of the boy and the skateboard is 1.93 m/s.    

Explanation:

We can find the final velocity of the boy and the skateboard by conservation of linear momentum:

$P_{1} = P_{2}$

$m_{b}v_{i_{b}} + m_{s}v_{i_{s}} = m_{b}v_{f_{b}} + m_{s}v_{f_{s}}$  (1)

Where:

$m_{b}$: is the boy's mass = 55 kg

$m_{s}$: is the skateboard's mass = 2.0 kg

$v_{i_{b}}$: is the initial speed of the boy = 2.0 m/s

$v_{i_{s}}$: is the initial speed of the skateboard = 0

$v_{f_{b}}$: is the final speed of the boy =?

$v_{f_{s}}$: is the final speed of the skateboard =?

Since the boy and the skateboard will have the same final velocity by entering the above values into equation (1) we have:

$55 kg*2.0 m/s + 0 = v_{f}(55 kg + 2.0 kg)$

$v_{f} = 1.93 m/s$

Therefore, the final velocity of the boy and the skateboard is 1.93 m/s.    

I hope it helps you!                                

Answer 2

The final velocity of the boy and the skateboard is 1.93m/s

According to the law of conservation of momentum, the momentum of the bodies before the collision is equal to the momentum of the bodies after the collision.

Mathematically, $m_1u_1+m_2u_2=(m_1+m_2)v$

$m_1 \ and \ m_2$ are the masses of the object

$u_1 \ and \ u_2$ are the velocities of the object

v is the final velocity

Given the following parameters

m₁ = 55kg

u₁ = 2.0m/s

m₂ = 2.0kg

u₂ = 0m/s

Required

Final velocity "v"

Substitute the given parameters into the formula to have:

$55(2)+2(0)=(55+2)v\\110=57v\\v=\frac{110}{57}\\v= 1.93m/s$

Hence the final velocity of the boy and the skateboard is 1.93m/s

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