A steel ball of mass 0.260 kg is fastened to a cord that is 33.0 cm long and fix
ID: 2111554 • Letter: A
Question
A steel ball of mass 0.260 kg is fastened to a cord that is 33.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal, as shown in Fig. 9-65. At the bottom of its path, the ball strikes a 4.20 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.
A steel ball of mass 0.260 kg is fastened to a cord that is 33.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal, as shown in Fig. 9-65. At the bottom of its path, the ball strikes a 4.20 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.Explanation / Answer
Find the speed of the ball at the point it hits the block using energy conservation: 0.5mv² = mgh v = v[2gh] = v[2(9.81m/s²)(0.500m)] = 3.13m/s Now use the law of conservation of momentum to find the speed of the ball and the block after the collision: m1v1(i) + m2v2(i) = m1v1(f) + m2v2(f) (0.600kg)(3.13m/s) + 0 = m1v1(f) + m2v2(f) 1.88kg•m/s = m1v1(f) + m2v2(f)---------------------------->(1) Since we have two unknowns, we need another equation: v1(i) - v2(i) = -[v1(f) - v2(f)] 3.13m/s - 0 = v2(f) - v1(f) Solving this for v2(f): v2(f) = 3.13m/s + v1(f) --------------------------->(2) Plugging (2) into (1) eliminates v2(f) so we can find v1(f): 1.88kg•m/s = m1v1(f) + m2[3.13m/s + v1(f)] 1.88kg•m/s = (0.600kg)v1(f) + (2.00kg)[ 3.13m/s + v1(f)] 1.88kg•m/s = (0.600kg)v1(f) + 6.26kg•m/s + (2.00kg)v1(f) (2.60kg)v1(f) = -4.38kg•m/s v1(f) = -1.68m/s----------------------->answer to (a), minus as it reverses direction The speed of the block can be found from equation (2): v2(f) = 3.13m/s + v1(f) = 3.13m/s - 1.68m/s = 1.45m/s------------------------------>an… to (b)
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