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Problem 16.40 The 1-kg ball is given a horizontal velocity of 1.2 m/s at A. Photographic measurements indicate that b = 1.2 m, h = 1.3 m, and the duration of the bounce at B is 0.1 s. What are the components of the average impulsive force exerted on the ball by the floor at B? y A 2m h Solution: The impulse is x B t2 b F dt = Fave (t2 − t1 ) = m(v2 − v1 ). t1 Subtract the weight of the ball from the average impulsive force: The velocities are determined from the path (from Newton’s second √ law for free fall): v1 = 1.2i − 2gyj = 1.2i − 6.26j (m/s). The verti√ cal velocity after the bounce at B is 2gh = 5.05 m/s. The time of flight after the bounce at B is twice the time required to fall a distance h, from which Fimp = Fave − (−mgj) = −0.345i + 123j (N) y 2h = 1.03 s. t =2 g 2m The horizontal velocity after the bounce at B is h B b / t = 1.17 m/s, ovrbt √ v2 = 1.2i + x b 2 ghj = 1.17i + 5.05j (m/s). From which t2 F dt = Fave (t2 − t1 ) = −0.0345i + 11.31j N-s. t1 Fave = 1 (−0.0345i + 11.31j) = −0.345i + 113.1j (N). 0.1 Problem 16.41 At time t = 0, the two masses are released from rest on the smooth surface with the spring stretched. Show that at any later time t, the velocities of the masses are related by k mB mA mA vA + mB vB = 0. Strategy: Write the principle of impulse and momentum for each mass. Solution: The force on mass A is equal and opposite to that on mass B. Hence, we can write FA + FB = 0 and t FA dt + 0 0 t FB dt = t 0 dt = 0. 0 From the principle of impulse and momentum, we can write t v FA dt = mA 0 dv = mA vA . 0 Similarly, we can write 0 t FB dt = mB v k mA dv = mB vB . 0 Substituting into the equation above, we get mA vA + mB vB = 0, as was required. mB