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Novosadov [1.4K]

3 years ago

8

A 4 kg body is traveling at 3 m/s with no external force acting on it. At a certain instant an internal explosion occurs, splitt

ing the body into two chunks of 2 kg mass each. The explosion gives the chunks an additional 18 J of kinetic energy. Neither chunk leaves the line of original motion. Determine the speed and direction of motion of each of the chunks after the explosion.

1 answer:

Ann [662]3 years ago

5 0

Answer

given,

mass of body = 4 Kg

travelling at speed = 3 m/s

exploded into two fragments of equal mass i.e. = 2 Kg

energy after collision increases by 18 J

original kinetic energy

$KE = \dfrac{1}{2}MV^2$

$KE = \dfrac{1}{2}\times 4 \times 3^2$

KE = 18 J

energy increases after collision = 18 J

total energy = 36 J

now,

$\dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2}m_2v_2^2 = 36$

$\dfrac{1}{2}v_1^2 + \dfrac{1}{2}v_2^2 = 36$

$v_1^2+v_2^2 = 72$

using conservation of momentum

m₁v₁ + m₂v₂ = MV

2 v₁ + 2 v₂ = 4 x 3

v₁ + v₂ = 6-------------(1)

squaring both side in the above equation

v₁² + v₂² + 2v₁v₂ = 36

72 + 2v₁v₂ = 36

v₁v₂ = -18-------(2)

putting value of equation (1) in (2)

v₁ (-18 - v₁) = 6

-18 v₁ - v₁² = 6

v₁² + 18 v₁ + 6 = 0

on solving the above equation we get

v₁ = -0.34 m/s

putting value of v₁ in equation (1)

v₂ = 6 + 0.34

v₂ = 6.34 m/s

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Complete question is;

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