A 12.0 g wad of sticky clay is hurled horizontally at a 110 g wooden block initi

Question

A 12.0 g wad of sticky clay is hurled horizontally at a 110 g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between the block and the surface is 0.850, what was the speed of the clay immediately before impact?

xxxxx104.328 is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s xxxxx

Explanation / Answer

Given that the mass of clay is m = 12.0 g = 0.012 kg Mass of wooden block is M = 110 g = 0.110 kg Distance traveled by the block and clay before coming to rest is S = 7.50m The coefficient of friction between the block and the surface is n = 0.850 ------------------------------------------------------------------------------------------- Let u be the speed of clay before collision Let v be the speed of the block and the clay. Apply conservation of momentum before and after collision mu + M(0) = (m + M)v -----------------(1) ------------------------------------------------------------------------------------------ Since the two bodies moving on the rough horizontal surfce, the acceleration of the two bodies is a = - ng ----------(2) (Since there is no external force other than the frictional force) but from equation of motion acceleration is a = (V^2 - v^2) / 2S a = (0 - v^2) / 2S ----------- (3) From equations (1) and (2) we get v^2 / 2S = ng v = (2ngS)^1/2 = (2*0.85*9.8m/s^2*7.50)^1/2 = 11.17 m/s Substitude this value in eqation (1) we get u = (m + M)(11.17 m/s) / m u = (1 + M/m)(11.17 m/s) u = 113.56 m/s

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