Modified Newtons cradle and the Conservation of Momentum

[shyguy79]

Homework Statement

2 balls hang from a modified Newtons Cradle, on of mass 80g, the other of mass 200g. Assuming gravity is 9.81ms^-2 ball A is pulled out so it reaches a height of 10cm then released. After impact ball 2 moves at a speed of 0.80ms^-1... calculate the velocity of sphere A immediately after impact.

A total of 9 marks

Homework Equations

E grav = mgh
E trans = 0.5mv^2
The law of conservation of momentum

The Attempt at a Solution

(a) m1 = 0.08kg m2 = 0.2kg v2 = 0.8ms-1 v1 = ? height = 0.1m

Neglecting air resistance when ball A is pulled up to a height of 0.10m where its gravitational potential energy is at a maximum:

Egrav = mgh = 0.08kg x 9.81ms-1 x 0.1m = 0.078 Joules

The ball is then released and Egrav is transferred to kinetic energy Etrans fully when the ball reaches the bottom of the swing so Egrav will be at a minimum and Etrans will be at a maximum.

So if Etrans = 0.5mv^2 then assuming there is no energy loss then Etrans = 0.078J

Therefore rearranging Etrans for v then the velocity at the point of impact is:

v1 = 1.4ms-1 at the point of impact.

The kinetic energy of m2 after being struck is Etrans = 0.5 = 0.5 x 0.2 x 0.82 = 0.064 J

So the difference in energy would be 0.078J - 0.064J = 0.014J

so the velocity of m1 after the collision v = √(0.014J) / (0.5 x 0.08kg) = 0.6ms-1

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My answer just seems a long winded way of doing it but for 9 marks I can't see where I've gone wrong if I have. It feels wrong because v1 of m1 is just the difference between the initial velocity of the m1 and the final velocity of m2...

Any pointers would be greatly appreciated.

 

[PeterO]

Homework Statement

2 balls hang from a modified Newtons Cradle, on of mass 80g, the other of mass 200g. Assuming gravity is 9.81ms^-2 ball A is pulled out so it reaches a height of 10cm then released. After impact ball 2 moves at a speed of 0.80ms^-1... calculate the velocity of sphere A immediately after impact.

A total of 9 marks

Homework Equations

E grav = mgh
E trans = 0.5mv^2
The law of conservation of momentum

The Attempt at a Solution

(a) m1 = 0.08kg m2 = 0.2kg v2 = 0.8ms-1 v1 = ? height = 0.1m

Neglecting air resistance when ball A is pulled up to a height of 0.10m where its gravitational potential energy is at a maximum:

Egrav = mgh = 0.08kg x 9.81ms-1 x 0.1m = 0.078 Joules

The ball is then released and Egrav is transferred to kinetic energy Etrans fully when the ball reaches the bottom of the swing so Egrav will be at a minimum and Etrans will be at a maximum.

So if Etrans = 0.5mv^2 then assuming there is no energy loss then Etrans = 0.078J

Therefore rearranging Etrans for v then the velocity at the point of impact is:

v1 = 1.4ms-1 at the point of impact.

The kinetic energy of m2 after being struck is Etrans = 0.5 = 0.5 x 0.2 x 0.82 = 0.064 J

So the difference in energy would be 0.078J - 0.064J = 0.014J

so the velocity of m1 after the collision v = √(0.014J) / (0.5 x 0.08kg) = 0.6ms-1

----------------
My answer just seems a long winded way of doing it but for 9 marks I can't see where I've gone wrong if I have. It feels wrong because v1 of m1 is just the difference between the initial velocity of the m1 and the final velocity of m2...

Any pointers would be greatly appreciated.

With a traditional Newtons cradle we usually get an elastic collision so this may be correct, but we can be certain that momentum is conserved so you could double check that way.

 

[shyguy79]

thanks for the reply. Using the conservation of momentum and p1 = m1v1 should equal p2 = m2v2 in an elastic collision...

P1 = m1v1 = 0.08kg * 1.4ms-1 = 0.112
P2 = m2v2 = 0.2kg * 0.8ms-1 = 0.16

If you take the difference P2-P1 = 0.048

Rearrange P=mv for v: v=P/m = 0.048/0.08 = 0.6ms-1

So again the same answer?