Answer:
The answer is in the explanation
Explanation:
A)
i) The blocks will come to rest when all their initial kinetic energy is dissipated by the friction force acting on them. Since block A has higher initial kinetic energy, on account of having larger mass, therefore one can argue that block A will go farther befoe coming to rest.
ii) The force on friction acting on the blocks is proportional to their mass, since mass of block B is less than block A, the force of friction acting on block B is also less. Hence, one might argue that block B will go farther along the table before coming to rest.
B) The equation of motion for block A is
m_{A}\frac{\mathrm{d} v}{\mathrm{d} t} = -m_{A}g\nu_{s}\Rightarrow \frac{\mathrm{d} v}{\mathrm{d} t} = -\nu_{s}g \quad (1)
Here, \nu_{s} is the coefficient of friction between the block and the surface of the table. Equation (1) can be easily integrated to get
v(t) = C-\nu_{s}gt \quad (2)
Here, C is the constant of integration, which can be determined by using the initial condition
v(t=0) = v_{0}\Rightarrow C = v_{0} \quad (3)
Hence
v(t) = v_{0} - \nu_{s}gt \quad (4)
Block A will stop when its velocity will become zero,i.e
0 = v_{0}-\nu_{s}gT\Rightarrow T = \frac{v_{0}}{\nu_{s}g} \quad (5)
Going back to equation (4), we can write it as
\frac{\mathrm{d} x}{\mathrm{d} t} = v_{0}-\nu_{s}gt\Rightarrow x(t) = v_{0}t-\nu_{s}g\frac{t^{2}}{2}+D \quad (6)
Here, x(t) is the distance travelled by the block and D is again a constant of integration which can be determined by imposing the initial condition
x(t=0) = 0\Rightarrow D = 0 \quad (7)
The distance travelled by block A before stopping is
x(t=T) = v_{0}T-\nu_{s}g\frac{T^{2}}{2} = v_{0}\frac{v_{0}}{\nu_{s}g}-\nu_{s}g\frac{v_{0}^{2}}{2\nu_{s}^{2}g^{2}} = \frac{v_{0}^{2}}{2\nu_{s}g} \quad (8)
C) We can see that the expression for the distance travelled for block A is independent of its mass, therefore if we do the calculation for block B we will get the same result. Hence the reasoning for Student A and Student B are both correct, the effect of having larger initial energy due to larger mass is cancelled out by the effect of larger frictional force due to larger mass.
D)
i) The block A is moving in a circle of radius L+\frac{d}{2} , centered at the pivot, this is the distance of pivot from the center of mass of the block (assuming the block has uniform mass density). Because of circular motion there must be a centripetal force acting on the block in the radial direction, that must be provided by the tension in the string. Hence
T = \frac{m_{A}v^{2}}{L+\frac{d}{2}} \quad (9)
The speed of the block decreases with time due to friction, hence the speed of the block is maximum at the beginning of the motion, therfore the maximum tension is
T_{max} = \frac{m_{A}v_{0}^{2}}{L+\frac{d}{2}} \quad (10)
ii) The forces acting on the block are
a) Tension: Acting in the radially inwards direction, hence it is always perpendicular to the velocity of the block, therefore it does not change the speed of the block.
b) Friction: Acting tangentially, in the direction opposite to the velocity of the block at any given time, therefore it decreases the speed of the block.
The speed decreases linearly with time in the same manner as derived in part (C), using the expression for tension in part (D)(i) we can see that the tension in the string also decreases with time (in a quadratic manner to be specific).