Question

plz solve it quickly​

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

Answer:

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'

p′=p12+p22+2p1p2 cos 90∘\sf p' = \sqrt{{p}^{2}_1 + {p}^{2}_2 + 2{p}_1{p}_2 \ cos \ {90}^{ \circ} }p

′

=

p

1

2

+p

2

2

+2p

1

p

2

cos 90

∘

p′=62+62+0\sf p' = \sqrt{{6}^{2} + {6}^{2} + 0 }p

′

=

6

2

+6

2

+0

p′=2(62)\sf p' = \sqrt{2({6}^{2} )}p

′

=

2(6

2

)

p′=62 kgms−1\sf p' = 6 \sqrt{2} \ kgm {s}^{ - 1}p

′

=6

2

kgms

−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.