SPRING 2002 MIDTERM EXAMINATION II

PHYSICS 220A #85077

1. A 10.0kg block is released from point A in Figure. The track is frictionless except for the portion BC, of length 6.00m. The block travels down the track, hits a spring of force constant k = 2250N/m, and compresses it 0.300m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between surface BC and block.

Solution.

 

 

 Choose the initial position when the block is at A and the final position when the spring is fully compressed.

                   (K + Ug + Us)i  + WNC =  (K + Ug + Us)f.      

                    0 + mgyA +  0 + FfrBCcos180° = 0 + 0 + 1/2kx2; 

                   (10.0kg)(9.80m/s2) - Ffr(6.00m) = 1/2(2250N/m)(0.300m)2,

which gives Ffr = 32.1N.

Now consider the friction force, we have Ffr = mkFN.

From SF = ma for the block we have

                    FN - mg = 0.

So we have   Ffr = mkmg;

                     (32.1N) = (10.0kg)(9.80m/s2),

which gives mk = 0.328.        

2. A 3.0-kg steel ball strikes a wall with a speed of 10m/s at an angle of 60° with the surface. It bounces off with the same speed and angle (figure). If the ball is in contact with the wall for 0.20s, what is the average force exerted on the ball by the wall.            

Solution.

 

 We resolve the momentum into two components, and we have

                Dp = Ft 

      y-component:  Dpy = m(vyf - vyi) = m(vcos60°) - mvcos60° = 0.

      x-component:  Dpx = m(vyf - vyi) = m(-vsin60° - vsin60°) = -2vsin60°

                                    = -(2(3.0kg)(10m/s)(0.866) = -52kg·m/s

                F = Dp/t = Dpxi/t = -52kg·m/s/(0.20s) = -260iN.

 

3. How much work is done by the Moon's gravitational field as a 1000-kg meteor comes in from outer space and impacts on the Moon's surface? The mass of Moon is 7.36 x 1022kg, and the Moon's radius is 1.74 x 106m.

Solution.

We find the force between the Moon and the meteor from

                           F = (GmMoonm/r2)r.

Then the work done by the Moon gravitational field is

                          W = ò¥r(GmMoonm/r2)r·dr = ò¥1.74 x 10^6m[(6.67 x 10-11N·m2/kg2)(7.36 x 1022kg)(1000kg)/r2]dr

                              = 4.47 x 1018N·m2(-1/r)|¥1.74 x 10^6m

                              = 2.82 x 109J.

4. Three particles of mass m1,  m2, and m3 are located at the three corners of an equilateral triangle with 2a on a side. They are connected by massless cords as shown. Find the center of mass of the combination.

Solution.

 

 We choose the coordinate system as shown.

We find the center of mass of the combination from

                   xCM = (x1m1+ x2m2+ x3m3)/(m1+ m2 + m3)

                            = [(-a)m1 + (a)m2 + (0)m3]/(m1+ m2 + m3)

                            = (m2 - m1)a/(m1+ m2 + m3).

                     yCM = (y1m1+ y2m2+ y3m3)/(m1+ m2 + m3)

                             = [(0) + (0) + 2asin60°m3]/(m1+ m2 + m3)

                             = Ö3am3/(m1+ m2 + m3).

 

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