### Momentum

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• #### Message 1.

Posted by BitesizePhysicsTeacher (U15165682) on Saturday, 9th February 2013

Momentum is the product of the mass and velocity of a moving object, it has the symbol p, i.e. p = m v. and it is measured in kg m/s.

Momentum is important when moving objects interact - i.e. crash into each other.

In a collision between two objects (or an explosion, where things fly apart) if no other external forces are acting the momentum is always CONSERVED.

i.e. total momentum before collision = total momentum after collision

In Higher, collisions only ever happen in a straight line, and there are only ever two objects involved.

If we use the usual symbols for mass - m, initial velocity - u, and final velocity - v, and numbers for the two objects, then we can use the following equation -

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

In any question, we can find an unknown value by substituting the information given in the question and solving.

Example -

A golf club of mass 0.35 kg travelling at 15 m/s hits a stationary golf ball with a mass of 0.045 kg. If the golf ball travels away with a velocity of 20 m/s, what is the speed of the club after impact?

m₁ = 0.35 kg, u₁ = 15 m/s, v₁ = ?
m₂ = 0.045 kg, u₂ = 0 m/s, v₂ = 20 m/s

=> m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

=> 0.35 x 15 + 0.045 x 0 = 0.35 x v₁ + 0.045 x 20

=> 5.25 + 0 = 0.35 x v₁ + 0.9

=> v₁ = (5.25 - 0.9) ÷ 0.35

=> v₁ = 12.4 m/s

• #### Message 2

, in reply to message 1.

Posted by Honeybee (U15216393) on Sunday, 28th April 2013

A ball of mass 100g rolls down a curved curtain rail and collides with a stationary ball of mass 50g. After the collision both balls move forward in a horizontal direction and are projected off the end of the rail, with ball A travelling 0.75m and ball B travelling 1.5m. The height of the curtain rail is 1.25m

Calculate
1)the speed of A and B just after the collision
2) the speed of A just before the collision
3) the initial height of ball A

• #### Message 3

, in reply to message 2.

Posted by BitesizePhysicsTeacher (U15165682) on Monday, 29th April 2013

Thanks for your question, it's quite a fiddly one, but if we take it slowly we should be fine.

1) To solve it we first need to find a time, and the only way we have to do this is using the height of the curtain rail and the fact that the balls are HORIZONTALLY projected.

This means that in the vertical direction they have initial velocity of 0m/s, so taking gravitational acceleration as 10 and using s for the vertical displacement (height), we get -

s = ut + ½at²
1.25 = (0xt) + (½x10xt²)

Rearranging, we get -

t² = 0.25

so

t = 0.5 s

This means we can use the HORIZONTAL displacements to find the horizontal velocities after collision, because the horizontal velocity is unchanged for projectiles.

for A, v = s / t = 0.75 / 0.5 = 1.5 m/s

for B, v = s / t = 1.5 / 0.5 = 3.0 m/s

2) To find the speed of A before the collision, we use the principle of conservation of momentum - total momentum before collision = total momentum after collision

3) To find the starting height for ball A, we use the principle of the conservation of energy - gravitational potential energy lost = kinetic energy gained

for A, Ek = ½mv² = ½ x 0.1 x 1.5² = 0.1125 J

so at height, h, Ep = mgh = 0.1125 J

therefore, h = Ep / mg = 0.1125 / (0.1x10) = 0.1125 m

Hope this helped.

BPT

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