Two particles moving in opposite direction collide inelastically. The velocity of the firstparticle before the collision is 6 m/s and the second particle is 12 m/s. After collision,both particle moves with a common velocity of 3 m/s in the direction of first particle.Find the ratioof the kinetic energy of the first particle to that of the second particle before the collision
Question
Two particles moving in opposite direction collide inelastically. The velocity of the firstparticle before the collision is 6 m/s and the second particle is 12 m/s. After collision,both particle moves with a common velocity of 3 m/s in the direction of first particle.Find the ratioof the kinetic energy of the first particle to that of the second particle before the collision
Solution
The kinetic energy of a particle is given by the formula KE = 1/2 * m * v^2, where m is the mass of the particle and v is its velocity.
Let's denote the mass of the first particle as m1 and the mass of the second particle as m2.
The kinetic energy of the first particle before the collision is KE1 = 1/2 * m1 * (6 m/s)^2 = 18 m1 Joules.
The kinetic energy of the second particle before the collision is KE2 = 1/2 * m2 * (12 m/s)^2 = 72 m2 Joules.
The ratio of the kinetic energy of the first particle to that of the second particle before the collision is therefore KE1/KE2 = 18 m1 / 72 m2 = m1 / 4m2.
This ratio depends on the masses of the particles, which are not given in the problem. If the masses are equal (m1 = m2), the ratio would be 1/4. If the mass of the first particle is four times the mass of the second particle (m1 = 4m2), the ratio would be 1.
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