Question

An object at rest suddenly explodes into three parts of equal mass. Two of the parts move away at right angles to each other and with equal speeds v. What is the velocity of the third part just after the explosion?

(a) Direction of vector 1 and magnitude 2v
(b) Direction of vector 2 and magnitude √2v
(c) Direction of vector 3 and magnitude

(d) Direction of vector 2 and magnitude

(e) Direction of vector 1 and magnitude

Answer 1

According to conservation of linear momentum, the initial momentum before the collision is equal to the final momentum after the collision.

Pi = Pf

 

Here, Pi is the initial momentum and Pf is the final momentum.

Here, initially the object is at rest. The initial momentum before explosion is,

Pi = 0

 

Here, the two parts of each mass are at right angles to each other with same speed. Therefore, the vertical component of momentum is zero.

 

Horizontal component of momentum is,

 

PHorizontal = mv cos45° + mv cos 45°

= 2mv cos 45°

= √2v

To make the horizontal component of the momentum to zero, the velocity of the third particle should be √2v in the direction of vector 2.

Therefore, the correct option is (b).

Therefore, the correct option is (b).