Question

A 7.0 kg mess kit sliding on a frictionless surface explodes into two 3.5 kg parts, one moving at 5.5 m/s, due north, and the other at 7.5 m/s, 30° north of east. What was the original speed of the mess kit? m/s

Answer 1

Answer:

The original speed of the mess kit is 5.65 m/s.

Explanation:

Given that,

Mass of kit =7.0 kg

Mass of first = 3.5 kg

Speed= 5.5 m/s

Speed of other particle part = 7.5 m/s

Angle = 30°

We need to calculate the momentum of first part

Using formula of momentum

$P=mv$

Put the value into the formula

$P_{1}=3.5\times5.5j$

$P_{1}=19.25j\ kgm/s$

We need to calculate the momentum of second part

Using formula of momentum

$P_{2}=3.5\times7.5(\cos30i+\sin30j)$

$P_{2}=26.25\times\dfrac{\sqrt{3}}{2}i+26.25\times\dfrac{1}{2}j$

$P_{2}=22.73i+13.13j$

We need to calculate the total momentum

$P=P_{1}+P_{2}$

$P=19.25j+22.73i+13.13j$

$P=22.73i+32.38j$

$P=\sqrt{(22.73)^2+(32.38)^2}$

$P=39.56\ kgm/s$

We need to calculate the original speed of the mess kit

Using formula of momentum

$P=MV$

$V=\dfrac{P}{M}$

Put the value into the formula

$V=\dfrac{39.56}{7.0}$

$V=5.65\ m/s$

Hence, The original speed of the mess kit is 5.65 m/s.