MIDTERM II

 1. Two masses are connected by a light string passing over a light frictionless pulley as shown in Figure 1. The 5.0-kg mass is released from rest. Using the law of conservation of energy, (a) determine the speed of the 3.0-kg mass just as the 5.0-kg mass hit the ground. (b) Find the maximum height to which the 3.0-kg mass rises.

Solution. (a) The energy before the 5.0-kg mass is released:

               Ei = 0 + m2gh = (5.0kg)(9.80m/s2)(4.0m) = 196J

              When the 5.0-kg mass hits the ground, the energy of the system is

              Ef  = 1/2(m1+ m2)v2 + m1gh= 1/2(3.0kg + 5.0kg)v2 + (3.0kg)(9.80m/s)(4.0m)

             By conservation of energy, we have Ei = Ef;

               196J = 1/2(8.0kg)v2 + 117.6J, which gives v = 4.43m/s.

               (b) When m1 reaches the maximum height, the speed of m1 is zero; we have

                  Ki + Ui = Kf + Uf

                  1/2m1v2 + m1gh = 0 + m1gH  

                 1/2(4.43m/s)2 + (9.80m/s)(4.0m) = (9.80m/s2)H , 

              which gives H = 5.00m                 

 

2. A 10.0-g bullet is stopped in a block of wood (m = 5.00kg). The speed of the bullet-plus-wood combination immediately after the collision is 0.600m/s. What was the original speed of the bullet?

Solution. From conservation of momentum, we have

              m1v1 + 0 = (m1 + m2)v'

              (10. x 10-3kg)v1 = (10 x 10-3kg + 5.00kg)(0.600m/s), which gives v = 3.01m/s.

 

              

 3. A circular hole of diameter a is cut out of a uniform square of sheet metal having sides 2a, as in Figure 2. Where is the center of mass for the remaining portion?

Solution. We treat the hole as a negative mass. From the diagram and symmetry,  we have

               x1 = a, x2 = 1.5a;    y1 = a,  y2 = 1.5a;

              Assume the density of the metal is s,

              xCM = [(2a)2s(a) - p(a/2)2s(1.5a)]/[(2a)2s - p(a/2)2s] = [a(32 - 3p)]/(16- p)

              Similarly, we have

              yCM = [a(32 - 3p)]/(16 - p)

 

Fig 2
 4. A 5.00-g bullet moving with an initial speed of 400m/s is fired into and passes through a 1.00-kg block, as in Figure 3. The block, initially at rest on a frictionless surface, is connected to a spring of force constant 900N/m. If the block moves 5.00cm to the right after impact, find (a) the speed at which the bullet emerges from the block and (b) the energy lost in the collision.

solution. We find the speed of the block immediately after the impact v2' first.

              From the conservation of energy, we have

                       Kblock  = Uspring

                       1/2mblockv2'2 = 1/2kx2

                                (1.00kg)v2'2 = (900N/m)(5.00 x 10-2m)2, which gives v2' = 1.5m/s

             From conservation of momentum, we have

                      mbulletv1 + 0 = mbulletv1' + mblockv2'

                      (5.00 x 10-3kg)(400m/s) = (5.00 x 10-3kg)v1' + (1.00kg)(1.5m/s), which gives

                         v1' = 100m/s.

             The initial energy of the system is the kinetic energy of the bullet:

                       Ei = Ki = 1/2mv12 = 1/2(5.00 x 10-3kg)(400m/s)2 = 400J

            The final energy of the system is the spring potential energy plus the kinetic energy of the bullet:

                       Ef = Uspring + 1/2mbulletv1'2 = 1/2kx2 + 1/2mbulletv1'2

                           =  1/2(900N/m)(0.05m)2 + 1/2(5.00 x 10-3kg)(100m/s)2 

                           = 1.125J + 25J = 26.125J

            The energy lost is

                     DE = Ei - Ef = 374J.

                       

                      

Fig 3