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 2620 N for 0.110 s. One piece of mass m1 = 2.80 kg travels backward at a velocity of 10.5 m/s and an angle of = 32.0

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 2620 N for 0.110 s. One piece of mass m1 = 2.80 kg travels backward at a velocity of 10.5 m/s and an angle of = 32.0degree above the horizontal. A second piece of mass m2 = 1.80 kg travels at a velocity of 8.80 m/s and an angle of 28.00degree below the horizontal. What is the velocity of the third piece? What is the direction of the third piece?

Explanation / Answer

impulse = Froce x time = change in momentum
2620 x 0.110 = 2.80 x (10.5cos32i + 10.5sin32j) + 1.80 x (8.80cos28i - 8.80sin28j) + (6.80 - 2.80 -1.80)v - (-6.80 x 22i)
288.2 = 188.52 i +8.14j + 2.2v
ball was in horizontal direction intially
so vj = 8.14/2.2 = 3.7j
vi =45.31 i

A_) velocity = 45.31i + 3.7j
magnitude = 45.46 m/s

b) direction = tan-1(3.7/45.31) = 4.67 degrees above

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