A 4.9 kg steel ball and 5 m cord ofnegligi- ble mass make up a simple pendulumth

Question

A 4.9 kg steel ball and 5 m cord ofnegligi-

ble mass make up a simple pendulumthat

can pivot without friction about the pointO.

This pendulum is released from rest in ahor-

izontal position and when the ball is atits

lowest point it strikes a 1.4 kg blocksitting

at rest on a shelf. Assume that thecollision

is perfectly elastic and take the coefficientof

friction between the block and shelf to be0.27.

The acceleration of gravity is 9.81m/s2.

= 0.27

5 m = height

1.4 kg= mass of block

4.9 kg = mass of ball

How far does the block move beforecoming

to rest?

Answer in units of m.

Explanation / Answer

A perfectly inelastic collision is defined as one in which noenergy is lost. Thus, we can use energy to solve this problem. The formula for the initial potential energy of the pendulumis: U = mgh Since the pendulum is held out to the horizontal, 5 m above thelowest point, h = 5 m.g = 9.81 m/s2, as stated, and m is4.9 kg, since the string is considered massless. Thus, the system will have a constant energy of U = mgh = (5m)(9.81 m/s2)(4.9 kg) = 240.345 J In order to set up the simultaneous equations for momentum, we needto know the velocity of the ball the instant before impact. Thesimple equation for the velocity of a falling object with no airresistance is v = 2gh. It's easy to see where this comesfrom by simplifying mgh = 1/2mv2. Regardless, v = 2(5 m)(9.81 m/s2) = 9.9 m/s Now we must set up the simultaneous equations for thecollision. m0v0 = m1v1 +m2v2 1/2m0v02 =1/2m1v12 +1/2m2v22 Where m0 and m1 are the mass of the pendulum, m2 is the mass of theblock, v0 is the velocity of the pendulum before impact, v1 is thevelocity of the pendulum after impact, and v2 is the velocity ofthe block after impact. Substitute and simplify: (4.9 kg)(9.9 m/s) = (4.9 kg)v1 + (1.4kg)v2(4.9 kg)(9.9 m/s)2 =(4.9kg)v12 + (1.4kg)v22 We need to solve this system of equations for v2, thespeed of the block after impact. We will solve the first equationfor v1 and substitute it into the second. 4.9v1 = 48.51 m/s - 1.4v2 v1 = 9.9 m/s - 0.2857v2 Substitute: (4.9 kg)(9.9 m/s)2 = (4.9kg)(9.9 m/s -0.2857v2)2 + (1.4kg)v22 Solving for v2 yields v2 = 15.4 m/s. Now all that remains is to find the displacement. For a constantforce, W = Fd. The block will stop when KE = W, so we need to findthe KE of the block just after impact: KE = 1/2(1.4 kg)(15.4 m/s)2 = 166 J Now we need to find F, the constant friction force: F = Fn = (0.27)(9.81 m/s2)(1.4 kg) =3.078 N Now, solving for d: 166 J = 3.078 N * d d = 44.77 m

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