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:
[tex] P_{1} = P_{2} [/tex]
[tex] m_{b}v_{i_{b}} + m_{s}v_{i_{s}} = m_{b}v_{f_{b}} + m_{s}v_{f_{s}} [/tex] (1)
Where:
[tex]m_{b}[/tex]: is the boy's mass = 55 kg
[tex]m_{s}[/tex]: is the skateboard's mass = 2.0 kg
[tex]v_{i_{b}}[/tex]: is the initial speed of the boy = 2.0 m/s
[tex]v_{i_{s}}[/tex]: is the initial speed of the skateboard = 0
[tex]v_{f_{b}}[/tex]: is the final speed of the boy =?
[tex]v_{f_{s}}[/tex]: 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:
[tex] 55 kg*2.0 m/s + 0 = v_{f}(55 kg + 2.0 kg) [/tex]
[tex] v_{f} = 1.93 m/s [/tex]
Therefore, the final velocity of the boy and the skateboard is 1.93 m/s.
I hope it helps you!
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, [tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
[tex]m_1 \ and \ m_2[/tex] are the masses of the object
[tex]u_1 \ and \ u_2[/tex] 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:
[tex]55(2)+2(0)=(55+2)v\\110=57v\\v=\frac{110}{57}\\v= 1.93m/s[/tex]
Hence the final velocity of the boy and the skateboard is 1.93m/s
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