Question

23. The object at rest suddenly explodes into three
parts with the mass ratio 2:1:1. The parts of
equal masses move at right angles to each
other with equal speed 'y'.the speed of the
third part after explosion will be
​

Answer 1

Dear Student,

◆ Answer -

v3 = -y/√2

● Explanation -

# Given -

m1 = m2 = m

|v1| = |v2| = y

m3 = 2m

# Solution -

Let M be mass of object. Object is initially at rest, therefore initial momentum will be zero.

Pi = Mu

Pi = M × 0

Pi = 0

Later when bomb exploded, final momentum of the system is -

Pf = m1v1 + m2v2 + m3v3

Pf = mv1 + mv2 + 2mv3

Pf = m(v1+v2) + 2mv3

In terms of magnitude,

Pf = m√(y²+y²+2y.y.cos90°) + 2mv3

Pf = √2my + 2mv3

From law of conservation of momentum,

Pi = Pf

0 = √2my + 2mv3

2mv3 = -√2my

v3 = -y/√2

Therefore, the speed of the

third part after explosion will be -y/√2.

Thanks dear...

Answer 2

Answer:

Explanation: here in the question they told that part of equal masses moves at right angles.. by applying coordinate system we get the answer quickly...