Question

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?

Answer 1

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 $=1.82\ ms^{-1}$