Question 1: In (Figure 1) , what value of F max gives an impulse of 4.8 Ns ? Que

Question

Question 1:

In (Figure 1) , what value of F max gives an impulse of 4.8 Ns ?

Question 2:

A package of mass m is released from rest at a warehouse loading dock and slides down the h = 2.4 m - high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m , from the bottom of the chute. (Figure 1)

A) Suppose the packages stick together. What is their common speed after the collision?

B) Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

Explanation / Answer

Queation 1) Impulse=Force* time

F_{max}=Impulse/time=4.8 /4 *10^{-3}=1.2kN

Queation 2)

A) By conservation of energy

mgh=(1/2)mv1^2

v1=sqrt{2gh}=sqrt{2*9.8*2.4}=6.858 m/s

By conservation of momentum

mv1 =(m+2m)Vf

Vf =v1 /3=6.858/3=2.286 m/s

b) Let the velocity of the m is v1 and the velocity of the 2m is v2 after the collision,by the energy conservation:-
=>v1 - v2 = u2 - u1
=>v1 - v2 = -6.858 -------------(i)
BY the law of momentum conservation:-
=>m1u1 + m2u2 = m1v1 + m2v2
=>m x 6.858 + 0 = mv1 + 2mv2
=>v1+2v2 = 6.858 ---------------(ii)
By 2 x (i) + (ii):-
=>3v1 = -6.858
=>v1 = -2.286 m/s
Thus again by the law of energy conservation,Let the package of mass m gain h meter due the velocity of 2.286 m/s
=>PE(top) = KE(bottom)
=>mgh = 1/2mv1^2
=>h = v1^2/2g
=>h = (2.286)^2/(2 x 9.8)
=>h = 0.267m

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