A steel ball of mass 0.50 kg is fastened to a cord 70.0 cm long and fixed at the

Question

A steel ball of mass 0.50 kg is fastened to a cord 70.0 cm long and fixed at the far end, and is released when the cord is horizontal. At the bottom of its path, the ball strikes a 2.50 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. (c) Find the angle from the horizontal the ball will swing up to in its back-swing. I came up with a) as v = sqrt of (2gl) part b) I got 0.74 m/s I can't seem to get part c, or even know how to approach it! If you could go through it step by step I would greatly appreciate it!

Explanation / Answer

First step is to determine the speed of the 0.500 ball just before impact with the 2.5 kg ball. To do this, apply conservation of energy Potential energy = Kinetic energy mgh = (1/2)mV^2 where m = 0.5 g = acceleration due to gravity = 9.8 m/sec^2 (constant) h = 70 cm = 0.70 m V = velocity of ball just before impact Substituting values, (0.5)(9.8)(.7) = (1/2)(0.5)V^2 V^2 = 13.72 V = 3.70 m/sesc. Applying conservation of momentum, Momentum before impact = Momentum after impact (.500)(3.7) + (2.5)(0) = 0.500(V1) + 2.5(V2) where V1 = velocity of the 0.500kg steel ball ater impact V2 = velocity of the 2.5kg steel block after impact Simplifying, 1.85 = 0.5V1 + 2.5V2 V1 = (1.85 - 2.5V2)/0.51 NOTE that, by definition, an elastic collision occurs when the kinetic energy of the system before impact is the same as the kinetic energy of the system after impact. Hence, KE before impact = (1/2)(0.5)(3.7)^2 + (1/2)(2.5)(0) KE before impact = 3.4225 KE after impact = (1/2)(0.5)V1^2 + (1/2)(2.5)V2^2 KE after impact = 0.25V1^2 + 1.25V2^2 Equating the two kinetic energy functions, 0.25V1^2 + 1.25V2^2 = 3.4225 Since V1 = (1.85 - 2.5V2)/0.51 0.25((1.85 - 2.5V2)/0.51)^2 + 1.25V2^2 = 3.4225 Simplifying the above 2.064V^2 - 6.1V2 - 0.1383 = 0 and using the quadratic formula, V = 2.97 m/sec. NOTE: I may have committed an error (or maybe errors due to carelessness) in the arithmetic process because my answer is different from what the answer is supposed to be (2.18 m/sec). But anyway, I would like to assure you that the analysis is the right one. 1. Use conservation of energy to determine the velocity of the 0.500 kg ball just before impact 2. Apply the conservation of momentum to determine the respective velocities of the steel ball and the block after impact. 3. Apply the principle that in an elastic collision, the kinetic energy of the system before impact is the same kinetic energy after impact

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.