Question

As shown in the Figure below, a light string that does not stretch changes from horizontal to vertical as it passes over the edge of a table. The string connects m_1, a 3.30 kg block, originally at rest on the horizontal table at a height 1.29 m above the floor, to m_2, a hanging 2 kg block originally a distance d = 0.980 m above the floor. Neither the surface of the table nor its edge exerts a force of kinetic friction. The blocks start to move from rest. The sliding block m_1 is projected horizontally after reaching the edge of the table. The hanging block m_2 stops without bouncing when it strikes the floor. Consider the two blocks plus the Earth as the system. 

 (a) Find the speed at which m_1 leaves the edge of the table. (Assume m_2 hits the ground before m_1 leaves the table.) 

 (b) Find the impact speed of m_1 on the floor. 

 (c) What is the shortest length of the string so that it does not go taut while m_1 is in flight? 

 (d) Is the energy of the system when it is released from rest equal to the energy of the system just before m_1 strikes the ground?

  Yes  No

  (e) Why or why not?

Answer 1

a.) After m2 gets lowered by a height of d = 0.98 m , the reduction in Potential energy = m2gd

PE = 2x 9.8 x 0.98 = 19.208 J

This is equal to the increase in the KInetic energy of the two blocks by the law of conservation of energy

So, 19.208 = 0.5 x (m1 + m2) x v²

19.208 = 0.5 x (3.3 + 2) x v²

v = 2.692267053 m/s

This is the velocity of both the blocks just before m2 hits the ground.

After m2 hits the ground, no more Potential energy is being reduced nor is any external force acting, to increase the Kinetic energy. So,

This will be the velocity with which m1 leaves the table.

b.) After leaving the table m1 falls for a height of h = 1.29 m

vertical component of velocity because of this fall be v'

v'² = 2gh = 2 x 9.8 x 1.29 = 25.284

v' = 5.028319799 m/s

So, the net velocity just before m1 hits the ground = ( v² + v'² )^1/2 = ( 2.692267053 ² + 5.028319799² )^1/2 = 5.703709485 m/s

c.) Time taken by m1 for the vertical fall t = v'/g = 5.028319799 / 9.8 = 0.513093857 seconds

We know that m1 had horizontal velocity v = 2.692267053 m/s

So, the horizontal distanca travelled by m1 in this time of fall = vt = 2.692267053 x 0.513093857 = 1.381385686 m

This is the shortest length required L = 1.381385686 m

d.) No,

e.) because m2 hits the ground before m1 and after hitting, the Kinetic energy of m2 gets dissipated to the environment as heat, sound etc .. (note that m2 stops after hitting the ground and the Kinetic energy of m2 is lost to surroundings) Because of this loss the Initial energy of the system will not be equal to the final energy of the system.