R-AEPHY-5

Momentum

• $p = mv$
• p is a vector quantity, $\therefore$ must put direction
• unit: $kg \ m \ s^{-1}$

Conserved quantity

• [insert conserved quality] can neither be created nor destroyed.
• Mass, momentum, force, energy

If a mass of 25.0kg is travelling at 4.00 ms^-1 east, what is its momentum

First answer: $100kg \ m \ s^{-1}$ East

This is wrong because for some god damn reason even though its 3 sig figs we don't know whether the 0's part of the significant figures or not. Hence, this is the correct answer $1 \times 10^{2} \ kg \ m \ s^{-1}$ East

What the hell dr waters

Conservation of momentum

• $\sum P$ before $= \sum P$ after
  • in other words, $m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1}+m_{2}v_{2}$
• e.g. two objects collide and remains separate

A billiard ball of mass 0.350 kg is travelling East at 4.51 $km h^{-1}$. It collides with a second billiard ball (same mass) which is initially stationary, but after the collision has velocity of 2.54 $kmh^{-1}$ east. What is the final velocity of the first ball?

Let east be positive

we know $m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1}+m_{2}v_{2}$. Thus $0.350 * 4.51 + 0.350 * 0 = 0.350 * 2.54+0.350 *v_{2}$ $0.350 * 4.51 = 0.350 * 2.54+0.350 *v_{2}$ $\frac{0.350 * 4.51 - 0.350 * 2.54}{0.350}= v_{2}$ $v_{2} = 1.97 km h^{-1}$

A truck of mass 1.50t is travelling west at 110 $kmh^{-1}$ when it has a head on collision with a car of 750 $kg$ travelling 74 $kmh^{-1}$ east. If the final velocity of the truck is 46.7 $km h^{-1}$ west, what is the final velocity of the car in $kmh^{-1}$?

Let west be positive we know $m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1}+m_{2}v_{2}$

$1500 * 110 + 750 * -74 = 1500 * 46.7 + 750 * v_{2}$ $= 52.6 kmh^{-1}$ $= 52.6 kmh^{-1}$ $= 14.6 ms^{-1}$

Important Equation

• $p = m(v-u)$
  • Note the $-u$
  • This applies when the vector directions aren't the same! i.e. when one vector is south, another is west

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