An m = 6.80-kg clay ball is thrown directly against a perpendicular brick wall a

Question

An m = 6.80-kg clay ball is thrown directly against a perpendicular brick wall at a velocity of 22.0 m/s and shatters into three pieces, which all fly backward, as shown in the figure. The wall exerts a force on the ball of 2690 N for 0.110 s. One piece of mass m1 = 2.80 kg travels backward at a velocity of 10.6 m/s and an angle of = 32.0° above the horizontal. A second piece of mass m2 = 1.80 kg travels at a velocity of 8.30 m/s and an angle of 28.0° below the horizontal.

a) What is the velocity of the third piece?


b) What is the direction of the third piece?

Explanation / Answer

Assuming the collision is perfect(the ball simply rebounds with the same momentum.) => The linear momentum is conserved . => 6.8 x 22 = (2.8 x cos 32 x 10.6) + (1.8 x cos 28 x 8.3) + m x v => m x v = 111.24 => As the sum of pieces weight equals the total weight of the ball m = (6.8 - 2.8 -1.8) = 2.2 kg Therefore the velocity of ball is , i.e, v = 111.24/2.2 = 50.56 m/s [(a) part answer] As there was no vertical momentum to the initial clay ball .. the piece will also not have any vertical momentum => 2.8 * 10.6 * sin32 - 1.8 * 8.3*sin 28 = 111.24 * sin (x) [x is angle with the negative vertical axis ] => 7.01 = 111.24 * sin(x) => x = 3.65 degrees below the horizontal . [(b) part answer]

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