Question

A 56-kg skater initially at rest throws a 5.0-kg medicine ball horizontally to the left. Suppose the ball is accelerated through a distance of 1.0 m before leaving the skater's hand at a speed of 7.0 m/s. Assume the skater and the ball to be point-like and the surface to be frictionless and ignore air resistance. Use a vertical y-axis with the positive direction pointing up and a horizontal x-axis with the positive direction pointing to the right. What will happen to the skater and the ball after the ball is thrown

Answer 1

Answer:

$a_2=24.5\ \text{m/s}^2$ towards right

$a_1=2.1875\ \text{m/s}^2$ towards left

Explanation:

$m_1$ = Mass of skater = 56 kg

$m_2$ = Mass of ball = 5 kg

v = Final velocity of ball = 7 m/s

u = Initial velocity of ball = 0

s = Distance the ball moved in the hand of the skater = 1 m

Moving left is considered and moving right is considered positive.

From kinematic equations of motion we have

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{7^2-0^2}{2\times 1}\\\Rightarrow a=24.5\ \text{m/s}^2$

So, the ball will move towards right with a magnitude of acceleration $a_2=24.5\ \text{m/s}^2$.

The force on the ball will be

$F_2=m_2a_2\\\Rightarrow F_2=5\times 24.5\\\Rightarrow F_2=122.5\ \text{N}$

The force on the ball is $122.5\ \text{N}$

The reaction force on the skater will be equal to the force on the ball but will have opposite direction.

$-F_1=F_2\\\Rightarrow -m_1a_1=F_2\\\Rightarrow a_1=-\dfrac{122.5}{56}\\\Rightarrow a_1=-2.1875\ \text{m/s}^2$

So, the skater will move towards left with a magnitude of acceleration $2.1875\ \text{m/s}^2$