Question

A student throws a 0.2 kg dart at a 1.2 kg dartboard suspended by a pair of chains from the ceiling. (A) If the dart was travelling horizontally at 9 m/s just before it hit the dartboard, how fast is the dartboard moving after the dart hits and sticks? (B) How high will the dartboard and dart now rise? (C) By calculations using your answer to part (A) and (B), is this collision elastic or inelastic?

Answer 1

The right concept to solve this problem is the conservation of the Moment.

By definition the conservation of the moment is given by,

$m_1u_1+m_2u_2= m_1v_1+m_2v_2$

Where,

$m_i =$ Mass of the objects/persons

$u_i =$ Initial Velocity of each one

$v_i =$ Final velocity of each one

PART A) For the first case the speed of the Dartboard is 0, and in the collision both objects maintain the same speed $v_1 = v_2$ therefore when replacing the data we have to

$m_1u_1+m_2u_2= m_1v_1+m_2v_2$

$(0.2*9)+0 =(0.2+1.2)(v_1+v_2)$

$(0.2*9)+0 =(0.2+1.2)V$

$V= 1.29 m/s$

Therefore the velocity of the dartboard moving after the dart hits and sticks is 1.29m/s

PART B) For conservation of potential and kinetic energy we have to

$KE = PE$

$\frac{1}{2}(m_T)v^2=m_Tgh$

Where

$m_T =$Total mass

Re-arrange for h,

$h=\frac{v^2}{2g}$

$h = \frac{1.29*1.29}{2*9.8}$

$h = 0.085m$

PART C) We have an inelastic collision. This understanding its definition in which kinetic energy is not conserved due to the action of internal friction.