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shusha [124]
3 years ago
10

A student m1 = 71 kg runs towards his skateboard, which has a mass m2 = 2.8 kg and is d = 2.75 m ahead of him. He begins at rest

and accelerates at a constant rate of a = 0.65 m/s2. When he reaches the skateboard he jumps on it. What is the velocity of the student and skateboard in meters per second?
Physics
1 answer:
diamong [38]3 years ago
7 0

Answer:

The velocity of the student and skateboard together =1.82\ ms^{-1}

Explanation:

Given:

Mass of student m_1=71\ kg

Mass of skateboard m_2=2.8\ kg

Distance between student and skateboard d=2.75\ m

Acceleration of student a=0.65\ ms^{-2}

Finding velocity v_1 of the student  before jumping on skateboard

Using equation of motion

v_1^2=v_0^2+2ad

here v_0 represents the initial velocity of the student which is =0 as he starts from rest.

So,

v_1^2=0^2+2(0.65)(2.75)

v_1^2=3.575

Taking square root both sides:

\sqrt{v_1^2}=\sqrt[1.7875}

∴ v_1=1.89

Finding velocity v of student and skateboard.

Using law of conservation of momentum.

m_1v_1+m_2v_2=(m_1+m_2)v

Where v_2 is initial velocity of skateboard which is =0 as it is at rest.

Plugging in values.

71(1.89)+(2.8)(0)=(71+2.8)\ v

134.19=73.8\ v

Dividing both sides by 73.8

\frac{134.19}{73.8}=\frac{73.8\ v}{73.8}

∴ v=1.82

The velocity of the student and skateboard together =1.82\ ms^{-1}

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