2. Refer to the diagram, which is an overhead view (and is not to scale). Two st

Question

2. Refer to the diagram, which is an overhead view (and is not to scale). Two stones, A and B, begin at rest on level, frictionless ice, as shown. The stones have unknown masses (that are not necessarily equal At time t = 0, a steady horizontal force of 80 N is applied to each stone. The direction of the force on stone A is due north (i.e. Z90°). The direction of the force on B is 240° (ie. 300° west of south). Both forces are removed at t = 4.00 s. At t= 9 s, the stones collide. The time graph here shows their interaction during the collision north east After the collision, stone A is moving directly south, and its speed is 1/3 of the speed it had just before the collision. Model the stones as particles and ignore all friction and air drag. FAB (N), a. Find FBA(magnitude and direction) during the collision b. Find the direction of stone B's motion after the collision. .1 02 0.3 04 05(ns)

Explanation / Answer

Each stone is accelerated for 4 s with force of 180 N

Let Ma and Mb the masses of each stone.

acceleration of A   = 180/Ma

starts from rest :

vel after 4 s   = 4*180/Ma = 720 /Ma

Force is withdrawn and the sotne moves with const. vel.

Momentum of A = vel * mass = 720 j   , let us take north as +ve y and east as +ve x

Pa = 720 j ;

vel of B   = 720/Mb

Momentum of B = 720 ( magnitude).

B makes 30 deg west of south

x-comp   = -720Cos(60) = -360

y -comp = -720Sin(60)   = -623.54

Pb = -360 i - 623 j ( vector notation)

Momentum is conserved during collision.

Total mpomentum before collision

Pa + Pb = 720 j -360 i - 623.5 j  

                = -360 i +92.5 j

Let Pa' and Pb' be the momentums after collision.

Pa' = -720 j /3 = -240 j ( A speed is 1/3 of its original speed)

moemntum of B after collision

                 = -360 i +92.5 j + 240 j

                 = -360 i + 332 j

direction of B Tan (b) = -332/360 =

b = -42.3 deg

B is headed 42.3 deg North of WEST.

a. Change in momentum of   = 720 + 240 = 960

Force F = dp/dt   = 960/ 0.5e-3 = 1920e+3 N

direction South    ( -j)

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