Q2b 2004

Q2b 2004 part 2

**Impacts & Collisions**

This topic is about spheres which collide with a barrier (impacts) or which collide with another sphere (collisions). There are four main types of question:

- Direct impacts
- Oblique impacts
- Direct collisions
- Oblique collisions

**Direct Impacts**

The equation which governs all impacts is *Newton’s Experimental Law (NEL):*

A particle (of mass 5 kg) falls vertically and hits the ground with speed, say, 40 m/s (i.e. its velocity vector is . If the coefficient of restitution *e* = 0.8, then

So the particle will rise with speed 32 m/s.

You must also know the definitions of impulse and kinetic energy. The symbol for impulse (a vector quantity, measured in Newton seconds) is *I* and is defined as :

In this case

The kinetic energy of a body (a scalar quantity measured in joules) is defined as *K = ½mv ^{2}*. Loss in kinetic energy =

*½mu*.

^{2}– ½mv^{2}Hence, in our case, the loss in kinetic energy =

*½(5)(40)*

^{2}^{ }–

*½(5)(32)*1440 joules

^{2}=**Oblique Impacts**

A particle of mass 4 kg hits horizontal ground with velocity m/s. The coefficient of restitution *e* = 0.5. What will the new velocity be after impact?

The key thing here to understand is that the i-speed will not change, but only the j-speed is subject to ** Newton’s Experimental Law**:

So the new speed will be m/s.

In this case,

It is worth noting that if the velocity of a particle is m/s, then its velocity *v* is determined by the equation *v ^{2} = x^{2} = y*

^{2 }(by Pythagoras’ Theorem) and hence, in this case, loss in K.E. =

*½mu*joules.

^{2}– ½mv^{2}= ½(4)[(3)^{2}+(-4)^{2}] – ½(4)[(3)^{2}+(2)^{2}] = 24Direct Collisions

The two equations which govern all collisions are the

** Law of Conservation of Momentum** (COM):

*m*

_{1}u_{1}+m_{2}u_{2}= m_{1}v_{1}+m_{2}v_{2 }

** Newton’s Experimental Law** (NEL):

You solve the simultaneous equations to find the unknown final velocities

*v*and

_{1}*v*.

_{2}**Oblique Collisions**

If we let the line of centres at impact be the i-axis, then the j-velocities will remain unchanged after the collision. Lets imagine this is the set-up:

Before collision (Mass) After Collision

First sphere *m*_{1 }

Second sphere *m _{2 }*

We want to find the unknowns *x* and *y.* We use the two simultaneous equations:

• COM: *m _{1}a + m_{2}c = m_{1}x +m_{2}y*

• NEL:

Solve these to get the answers.

There will be variations on this theme:

• If a sphere is at rest before impact, it will move along the i-axis after the collision.

• If the question states that before impact one of the spheres is at rest and that after impact, the spheres move at right angle to each other, then one sphere will move along the i-axis and the other along the j-axis.

• If you want to find the angle through which, for example, the second sphere above is deflected by the impact, follow these steps:

Find the slope of its path before =

Find the slope of its path after =

Find (the angle of deflection), using the formula .

**Common Mistakes**

• Forgetting to put the minus in front of *e* in NEL.

• Applying NEL to both directions, not just to the direction of impact.

• Confusing momentum and kinetic energy.

• Forgetting the definition of impulse, even though it is on Page 51 of The Mathematical Tables!