Question

A light string that does not stretch changes from horizontal to vertical as it passes over the edge of a table. The string connects m1, a 3.20 kg block, originally at rest on the horizontal table at a height 1.27 m above the floor, to m2, a hanging 3 kg block originally a distance d = 0.990 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 m1 is projected horizontally after reaching the edge of the table. The hanging block m2 stops without bouncing when it strikes the floor.

Answer 1

(a) Find the speed at which m1 leaves the edge of the table. (Assume m2 hits the ground before m1 leaves the table.) (b) Find the impact speed of m1 on the floor. (c) What is the shortest length of the string so that it does not go taut while m1 is in flight?

Answer 2

First, let's set up the equations of motion modeling the motion between the time the system is released from rest until the hanging mass hits the floor.

If we draw a free body diagram, we can see that for m_1\[\sum F_1 = m_1a = T\]and for m_2\[\sum F_2 = m_2a = T - m_2g\]T is the same for both masses, therefore\[m_1 a = m_2 a + m_2g\]Noting that both masses have the same acceleration\[a (m_1 - m_2) = m_2 g\]\[a = {m_2 g \over m_1 - m_2}\]

From constant acceleration kinematics, \[d = {1 \over 2} a t^2\]where \(d =0.990\) and a is the expression we found previously. We can solve this for t. Then\[v = at\]This will be the velocity of both masses after m2 hits the ground. Assuming that m1 is farther from the edge of the table than 0.990m, the velocity of m1 when it leaves the table will be equal to v as found from the last equation.

Now, we need to study m1 as we would in a projectile motion case. Once, it leaves the table, it has velocity in the x direction as found previously. It will maintain this velocity until it hits the ground. Once it leaves it table, gravity will begin accelerating it towards the ground. We can find the time it takes to hit the ground from\[d = {1 \over 2} g t^2\]where d is the height of the table, \(d = 1.27\)

The velocity of the object in the y-direction can be found as\[v_y = g t\]where t is the time taken to hit the ground, as found in the previous equation.

The speed when m1 hits the ground will be equal to \[s_g = \sqrt{v_x^2 + v_y^2}\]

To find the length of the spring, we need to determine if the range of the particle is greater than the height of the table. \[R = v_x t\]where t is the time of flight. If range is greater than the height of the table, we need to make the string equal to the range.