Question

A 4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the other at 6.8 m/s, 35° north of east. What is the original speed of the mess kit?

Answer 1

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Step-by-step explanation:

Let north represent positive y axis and east represent positive x axis.

Here momentum is conserved.

Let the initial velocity be v.

Initial momentum = 4.4 x v = 4.4v

Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s

Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s

Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s

We have

Initial momentum = Final momentum

4.4v = 12.364 i + 15.092 j

v =2.81 i + 3.43 j

Magnitude

`v=√(2.81^2+3.43^2)=4.43m/s`

Direction

`\theta =tan^(-1)\left ( (3.43)/(2.81)\right )=50.67^0`

50.67° north of east.

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.