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A particle of mass 5 kg, moving at 2 m/s, collides with a particle of mass 8 kg initially at rest. If the collision is elastic, find the velocity of each particle after the collision if the first particle is defected 50 o from its original direction of motion

User Nory
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Answer:

After the elastic collision, the first particle (5 kg) will be moving with a velocity of 2 m/s at an angle of 50 degrees to the left from its original direction of motion. The second particle (8 kg) will come to rest (velocity = 0 m/s).

Step-by-step explanation:

To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy for an elastic collision.

Conservation of momentum:

In an isolated system, the total momentum before the collision is equal to the total momentum after the collision.

Conservation of kinetic energy:

In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

Let's assume the initial velocity of the first particle (mass = 5 kg) is v1 and the final velocity after the collision is u1, and the initial velocity of the second particle (mass = 8 kg) is v2, and the final velocity after the collision is u2.

Step 1: Conservation of momentum

Initial momentum = Final momentum

m1 * v1 + m2 * v2 = m1 * u1 + m2 * u2

Step 2: Conservation of kinetic energy

Initial kinetic energy = Final kinetic energy

(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * m1 * u1^2 + (1/2) * m2 * u2^2

Given data:

m1 = 5 kg

v1 = 2 m/s

m2 = 8 kg

v2 = 0 m/s (initially at rest)

Step 3: Calculate the final velocities after the collision (u1 and u2).

Step 4: Calculate the angle of deflection for the first particle.

Let's solve the equations:

Step 1:

5 kg * 2 m/s + 8 kg * 0 m/s = 5 kg * u1 + 8 kg * u2

10 kg m/s = 5u1 + 8u2

Step 2:

(1/2) * 5 kg * (2 m/s)^2 + (1/2) * 8 kg * (0 m/s)^2 = (1/2) * 5 kg * u1^2 + (1/2) * 8 kg * u2^2

10 J = (1/2) * 5 kg * u1^2 + 0 J

Simplifying Step 2:

10 J = (1/2) * 5 kg * u1^2

u1^2 = (10 J) * 2 / 5 kg

u1^2 = 4 J/kg

u1 = √(4 J/kg)

u1 = 2 m/s (velocity of particle 1 after collision)

Step 1 (revisited):

10 kg m/s = 5 * 2 + 8 * u2

10 kg m/s - 10 kg = 8 * u2

u2 = (10 kg m/s - 10 kg) / 8

u2 = 0.0 m/s (velocity of particle 2 after collision)

Step 4:

The angle of deflection is given as 50 degrees from the original direction of motion for particle 1. Since the original direction was 2 m/s to the right, the deflection will be 50 degrees to the left.

Final answer:

After the elastic collision, the first particle (5 kg) will be moving with a velocity of 2 m/s at an angle of 50 degrees to the left from its original direction of motion. The second particle (8 kg) will come to rest (velocity = 0 m/s).

User A Stone Arachnid
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