A 0.0750kg ball is thrown at 25m/s towards a brick wall. (a) Determine the impul

Question

A 0.0750kg ball is thrown at 25m/s towards a brick wall. (a) Determine the impulse that the wall imparts to the ball when it hits and rebounds at 25m/s in the opposite direction. (b) Determine the impulse that the wall imparts to the ball when it hits and rebounds at an angle of 45.0 degrees. (c) If the ball thrown in part(b) contacts the wall for 0.0100s determine the magnitude and direction of the force that the wall exerts on the ball.

Explanation / Answer

a). Impulse = change in momentum => J = m*(Vf - Vi) => J = 0.075*[25-(-25)] => J = 3.75 N-s b). Vi = 25 (i) Vf = 25*cos45 (-i) + 25*sin45 (j) => Vf = - 17.68 i + 17.68 j J = m*[Vf - Vi] => J = 0.075*[(-17.68 i + 17.68 j) - (25 i)] => J = -3.201 i + 1.326 j magnitude of impulse = sqrt[(-3.201)^2 +1.326^2] = 12.005 N-s direction of impulse= 90 - tan^-1[1.326/3.201] = 67.5 degrees with the wall c). magnitude of force = J/t = 12.005/0.01 = 1200.5 N direction of force = 67.5 degrees with the wall

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