Question

An object of mass 4 kg is initially at rest on a frictionless ice rink. It suddenly explodes into three pieces (I have no idea why it happened!). One chunk of mass 1 kg slides across the ice with a velocity 1.2 ms^-1i and another chunk of mass 2kg slides across the ice with a velocity 0.8 m.s^-1j. Determine the velocity of the third chunk in magnitude angle form.

Answer 1

We have two events that occurred on different axes, the most convenient is to perform the operations on the two axes, and then look for the resulting force.

Our values are,

$m_1=1kg\\m_2 = 2kg\\v_1' = 1.2m/s\\v_1''= 0.8m/s$

PAR A) For the X axis, we apply momentum conservation, which is given by,

Total momentum before = Total momentum After

We start from rest, so in X the initial speeds are 0,

$m_1v_1+m_2v_2=m_1v_1'+m_2v_2'$

$0+0 = (1*1.2)+(1)v_2'$

$V_2' = -1.2m/s$

Now we apply for the conservation of the moment, it is part of the rest, so,

$m_1v_1+m_2v_2=m_1v_1''+m_2v_2''$

$0+0=2*0.8+1*v_2''$

$v_2'' = -1.6$

To find the total speed, we simply apply pitagoras,

$V = \sqrt{v_2'^2+v_2^2''}$

$V = \sqrt{1.2^2+1.6^2}$

$V = 2m/s$

PART B) The address is given by,

$tan\theta = \frac{V_2''}{V_2'}$

$\theta = tan^{-1} \frac{V_2''}{V_2'}$

$\theta = tan^{-1} \frac{1.6}{1.2}$

$\theta = 53.31\°$ (Below -x axis)