A ball with a mass of 0.605kg is initially at rest. It is struck by a second bal

Question

A ball with a mass of 0.605kg is initially at rest. It is struck by a second ball having a mass of 0.405kg , initially moving with a velocity of 0.255m/s toward the right along the x axis. After the collision, the 0.405kg ball has a velocity of 0.205m/s at an angle of 37.0? above the x  axis in the first quadrant. Both balls move on a frictionless, horizontal surface.

Part A) What is the magnitude of the velocity of the 0.605 kg ball after the collision?

Part B) What is the direction of the velocity of the 0.605 kg ball after the collision?

Part C) What is the change in the total kinetic energy of the two balls as a result of the collision?

Explanation / Answer

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

A ball with a mass of 0.585kg is initially at rest. It is struck by a second ball having a mass of 0.380 kg , initially moving with a velocity of 0.230m/s toward the right along the axis. After the collision, the 0.380kg ball has a velocity of 0.210m/s at an angle of 36.2degrees above the axis in the first quadrant. Both balls move on a frictionless, horizontal surface.

A. What is the magnitude of the velocity of the 0.585kg ball after the collision?

B. What is the direction of the velocity of the 0.585kg ball after the collision?

C. What is the change in the total kinetic energy of the two balls as a result of the collision?   

Answer

using law of conservation of momentum,

in the x - axis

m1u1 + m2u2 = m1v1cos? + m2v2cos?

0.38X0.23 = 0.38 X 0.21cos(36.2) + 0.585v2cos?

0.023 = 0.585v2cos?

v2cos? = 0.0393

in the y-axis,

0 = m1v1sin? - m2v2sin? [assuming that the other ball goes below the axis in the fourth quadrant]

0.585v2sin? = 0.38 X 0.21sin(36.2)

v2sin? = 0.0806

tan? = 0.0806/0.0393 = 2.051

? = 64 [below the x-axis]

A) the magnitude of the velocity of the 0.585kg ball after the collision = 0.0897m/s

B) the direction of the velocity of the 0.585kg ball after the collision = 64 degrees [southeast]

C) Kinetic energy before collision = 0.5 X 0.38 X 0.232 = 0.010051J

Kinetic energy after collision = 0.5 X 0.38 X 0.212 + 0.5 X 0.585 X 0.08972 = 0.010745J

Change = 6.94 X 10-4

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