Question

A bomb at rest explodes into three fragments of equal masses. Two fragments fly off at right angles to each other with velocity 9 m/s and 12 m/s. Calculate speed of third segment?

Answer 1

The initial momentum is 0

Initial momentum = final momentum

Final momentum:

Resultant momentum for the two is.

Mass is constant and depends not on the direction.

12² + 9² = 225

The resultant velocity = √225 = 15m/s

Since the mass is equal we have.

Momentum = 15m

Let the velocity of the third one be v

The final momentum is = mv + 15m

0 = mv + 15m

-15m = mv

v = - 15

The velocity of the third one is :

15 m/s

Answer 2

$\huge\textbf{QUESTION}$

$<I>{=>A bomb at rest explodes into three fragments of equal masses. Two fragments fly off at right angles to each other with velocity 9 m/s and 12 m/s. Calculate speed of third segment.}$

ANSWER

$<i>Initial momentum= 0$

We know that

Initial momentum = final momentum

Final momentum:

The Resultant momentum for the two fragments are =12² + 9² = 225

The resultant velocity = √225 = 15m/s

The mass is equal

so,we have.

Momentum = 15m

Let the velocity of the third segment be v

The final momentum is = mv + 15m

0 = mv + 15m

-15m = mv

v = - 15

Hence the velocity of the third segment is 15 m/s