Question

solve the problem...... ​

Answer 1

QUESTION:-

• A body of mass 1 kg breaks due to an explosion into three pieces of mass ratio 1 : 1 : 3. Two equal masses move at right angle to each other with a speed of 30m/s. What is the velocity of the heavier piece?

ANSWER:-

• 10√2 m/s at 135° with the direction of other two particles.

GIVEN:-

• A body of mass 1 kg breaks due to an explosion into three pieces of mass ratio 1 : 1 : 3.

• Two equal masses move at right angle to each other with a speed of 30m/s.

TO FIND:-

• The velocity of the heavier piece.

EXPLANATION:-

• Three masses m₁, m₂ and m₃ are in equilibrium.

• So the net momentum will be remain zero.

Total parts = 1 + 1 + 3

Total parts = 5

m₁ + m₂ + m₃ = 1 kg

[ 1/5 × (1) ] + [ 1/5 × (1) ] + [ 3/5 × (1) ] = 1

We can verify this,

1/5 + 1/5 + 3/5 = 1

(1 + 1 + 3)/5 = 1

5/5 = 1

1 = 1

m₁ = m₂ = 1/5 kg

m₃ = 3/5 kg

Momentum (p) = mv

p₁ = p₂ = m₁ v = 1/5 × (30) [ As m₁ = m₂ = 1/5 kg ]

p₁ = p₂ = 6 kgm/s

Angle between p₁ and p₂ is 90°

Let the resultant be p'

$\sf p' = \sqrt{{p}^{2}_1 + {p}^{2}_2 + 2{p}_1{p}_2 \ cos \ {90}^{ \circ} }$

$\sf p' = \sqrt{{6}^{2} + {6}^{2} + 0 }$

$\sf p' = \sqrt{2({6}^{2} )}$

$\sf p' = 6 \sqrt{2} \ kgm {s}^{ - 1}$

p' = p₃

p' = m₃ V

6√2 = 3/5 V

2√2 = 1/5 V

10√2 = V

v = 10√2 m/s

Angle between p' and p₁ = Angle between p' and p₂ [ Both are of same magnitude ]

Angle between p' and p₁ = 45°

Angle between p' and p₃ = 180°

Angle between p₁ and p₃ = Angle between p₂ and p₃ = 180° - 45° = 135°

• Hence the third piece moves with the velocity of 10√2 m/s making an angle of 135° with the direction of other two particles.

Answer 2

$\huge\red{Answer}$

Total mass of body m = 1 kg

So, mass of the first fragment m1 = 1/5×8 = 0.2kg

Mass of second fragment m2 = 1/5×8 = 0.2 kg

Mass of third fragment m3 = 3/5×1 = 0.6 kg

Velocity of masses m1 and m2 , v = 30m/s

Momentum of first fragment p1 = m1,v = 0.2× 30 = 6 kg m/s.

Similarly, p2 = m2,v = 0.2 × 30 = 6kg m/s.

Third fragment will fly in a direction opposite to the direction of resultant of p1 and p2. so that total momentum of the system after explosion remains zero.

Momentum of the third fragment

p3 = √p2/1 + √p2/2

= p3 = √6²+6² = 6√2 kg m/s

Thus speed of third fragment v3 =

p3/m3 = 6√2/0.6 = 10√2 m/s

Ans = x = 2