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 = √(v_2'^2+v_2^2'')`

`V = √(1.2^2+1.6^2)`

`V = 2m/s`

PART B) The address is given by,

`tan\theta = (V_2'')/(V_2')`

`\theta = tan^(-1) (V_2'')/(V_2')`

`\theta = tan^(-1) (1.6)/(1.2)`

`\theta = 53.31\°` (Below -x axis)