Question

A 5.0 kg stationary grenade explodes into three pieces. If a 0.5 kg piece moves at 35 m/s [E] and a 1.5 kg piece moves at 25 m/s [S 30° E], determine the velocity of the third piece?

Answer 1

I am using the trigonometry law to account for the final piece, well, as far as conservation of energy holds true .

Answer 2

The addition of all of these forces must equal zero

Answer 3

0.5kg/s*35m/s=17.5kgm/s
1.5kg/s*25m/s=37.5kgm/s
Now use the vector arrows to indicate for the hypotenuse

Answer 4

Use law of conservation of momentum
\[\vec p=\vec p_{1}+\vec p_{2}+\vec p_{3}\]
momentum before=momentum after
\[M\vec u=m_{1}\vec v_{1}+m_{2} \vec v_{2} +m_{3} \vec v_{3}\]
\[\vec v_{3}=\frac{1}{m_{3}}.(M \vec u-m_{1} \vec v_{1}-m_{2} \vec v_{2})\]

Answer 5

The triangle is imaginary, hypotenuse is pointing towards upper right.

Answer 6

We know that
\[\vec u=\vec 0\]
Therefore our equation becomes
\[\vec v_{3}=-\frac{m_{1} \vec v_{1}+m_{2} \vec v_{2}}{m_{3}}\]

Answer 7

Right, and final velocity obtained will be hence (17.5+37.5)N/3.0kg

Answer 8

Now I am only left to find the direction which satisfies the vector with cosine

Answer 9

or so I think

Answer 10

\[v_{3}=|\vec v_{3}|=|-\frac{m_{1}\vec v_{1}+m_{2} \vec v_{2}}{m_{3}}|=\frac{1}{m_{3}}.(m_{1}v_{1}+m_{2}v_{2})\]
\[v_{1}=|\vec v_{1}|=\sqrt{v_{1x}^2+v_{1y}^2}, v_{2}=|\vec v_{2}|=\sqrt{v_{2x}^2+v_{2y}^2}\]

Answer 11

Now I get it

Answer 12

But I think this assumes one dimensional momentum though

Answer 13

Two dimensions where two are canceling each other out becomes a question of finding the hypotenuse.

Answer 14

This is a general physics question dealing with the momentum?

Answer 15

Look