I know this is a multiple step problem and it build on each other. Bear with me.

Question

I know this is a multiple step problem and it build on each other. Bear with me.
This problem uses Newton's Cradle, with each mass being 3 kg.

A. Find the work done by gravity as the left ball goes from it's max height of 6 m to just before it hits the other balls.

B. Explain or show why if ball 1 in the arrangement is pulled back and let go, ball 5 bounces forwrd. If balls 1 nd 2 are pulled back and released, balls 4 and 5 bounce forward, and so on.

C. find the kinetic energy of the left ball just before it hits the other balls.

D. Find the potential energy of the left ball at its max height.

E. If the string that the ball is attached to is 100 m long, find the ngular speed of the ball just before it hits the other balls.

F. just before the ball hits, find the magnitude of the ball's angular acceleration. Also, illustrate the direction of the ball's linear acceleration.

G. Find the time to go from the location of max potential energy to minimum potential energy.

Please help me, I do rate very well.

Explanation / Answer

Part A)

W = PE = (mgh) = (3)(9.8)(6) = 176.4 J

Part B)

In Newton's Cradle, KE and Momentum must be conserved, so

.5mivi2 = .5mfvf2     and mivi = mfvf

Uning the momentum equation, mi = mfvf/vi

Sub that mi into the KE formula

.5(mfvf/vi)vi2 = .5mfvf2

Simplify

.5mfvf2 = .5mfvf2

This shows that the initial mass and final mass must be the same. Since all balls are the same value, if 1 hits initially, 1 flys out the other side. Same thing with 2 and 2,etc.

Part C)

KE = PE

KE = (m)(g)(h) = (3)(9.8)(6) = 176.4 J

Part D)

The PE is conserved, so it will still be 176.4 J

Part E)

Angular speed, = v/r

From the fact that we know KE = 176.4, that equals .5mv2

v = [(2(176.4)/3]

v = 10.8 m/s

= (10.8/100) = .108 rad/s

Part F)

The time required for the ball to swing down to its lowest point is found by

d = vot + .5at2

t = (2d/a) = [(2)(6)/(9.8)] = 1.11  sec

The using f = o + t, we can find

= (.108)/(1.11) = .098 rad/s2

Part G)

The time was found in the previous step

d = vot + .5at2

t = (2d/a) = [(2)(6)/(9.8)] = 1.11 sec

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---