Two rods, with masses MA and MB having a coefficient of restitution, e, move along a common line on a surface, figure 2. a) Find the general expression for the velocities of the two rods after impact. b) IfmA =2kg,mB =1kg,vB=3m/s,ande=0.65,find the value of the initial velocity vA of rod A for it to be at rest after the impact and the final velocity v’B of rod B. c) Find the percent decrease in kinetic energy which corresponds to the impact in part b
Two rods, with masses MA and MB having a coefficient of restitution, e, move along a common line on a surface, figure 2.
a) Find the general expression for the velocities of the two rods after impact.
b) IfmA =2kg,mB =1kg,vB=3m/s,ande=0.65,find the value of the initial velocity vA of rod A for it to be at rest after the impact and the final velocity v’B of rod B.
c) Find the percent decrease in kinetic energy which corresponds to the impact in part b
Let us assume that the velocity of bar A and B after impact be VA (dash) and VB (dash) respectively.
From the principle of conservation of momentum, Linear momentum before collision will be equal to final momentum after collision in the specific direction.
Balancing momentum in the x-direction as follows –
Writing the formula for the coefficient of restitution,
From equation (1) & (2),
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