A red stone with a mass of 2.5 kg slides on a smooth surface with a speed of 5.0

Question

A red stone with a mass of 2.5 kg slides on a smooth surface with a speed of 5.00 m/s along the x-axis. It has an elastic collision with a stationary blue stone which has a mass of 3.6 kg. After the collision, the red stone has a velocity of 4.00 m/s at an angle of 45.5 degrees relative to the x-axis. A) What is the initial kinetic energy? B) What is the velocity of the blue stone? C) What is vx and vy for the blue stone? D) What is the direction of the blue stone relative to the x-axis after the collision?

Thanks in advance for you help!

Explanation / Answer

Part A)

KE = .5mv2 = (.5)(2.5)(5)2 = 31.25 J

Part B)

By conservation of energy

KE after the collision is .5(2.5)(4)2 = 20 J

That means the blue stone has to make up for the difference

31.25 - 20 = 11.25 J

11.25 = (.5)(3.6)(v)2

v = 2.5 m/s

Part C)

Conservation of momentum in the x...

(2.5)(5) = 2.5(4)(cos 45) + (3.6)vx

vx= 1.51 m/s

Conservation of momentum in the y...

(2.5)(0) = 2.5(4)(cos 45) + (3.6)vy

vy = 1.96 m/s

Part D)

Direction from the tangent function

tan(angle) = 1.96/1.51

angle = 52.4o relative to the x axis on the opposite side as the red stone

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