Dual coding change of momentum

Rosencrantz (an anguished cry): CONSISTENCY IS ALL I ASK!

Tom Stoppard, Rosencrantz and Guildenstern Are Dead (1966)

I think that dual coding techniques can be extremely helpful in helping students understand the concept of change of momentum.

To engage our students’ physical intuitions, let’s consider a question like: Which would hurt more — being hit by a sandbag or being hit by a rubber ball?

Let’s assume that the sandbag and rubber ball have the same mass m and are travelling at the same initial velocity u. We choose ‘u‘ because it’s the initial velocity and we take ‘v‘ as the final velocity: a very subtle piece of dual coding that can reap rewards if applied consistently — pace Rosencrantz(!) — over a range of disparate examples.

To analyse this problem, let’s use the momentum version of Newton’s Second Law of Motion.

We will use the change = final – initial convention (‘Consistency is all I ask!’)). The initial momentum is pi and the final momentum is pf.

Now let’s work out the change in momentum in each case. We will assume that each item is dropped so that it impacts vertically on a horizontal surface. The velocity just before it hits is u so its initial momentum pi is given by pi = mu; its final velocity is v so its final momentum pf is given by pf = mv. The sandbag does not rebound, so its final velocity v is zero.

The rubber ball rebounds from the surface with a velocity v (we have shown that v < u so we are not assuming a perfectly elastic collision).

We will use the down-is-positive convention so that u is positive and the downward momentum pi are positive in both cases. However, the velocity v of the ball is negative so the momentum pf = mv is negative (upwards).

To add vectors, we simply put them ‘nose to tail’. However, in this case, we need to subtract the vectors, not add them. To do this, we use the operation pf + (-pi,). In other words, we put the vector pf nose to tail with minus pi, or with a vector pointing in the opposite direction to the original vector pi. These are shown in the table.

We can see that the change in momentum Δp is larger in the case of the rubber ball.

Applying Newton Second Law that force = change in momentum / change in time then (assuming the time of each interaction is the same) then we can conclude that the (upward) force exerted by the surface on the ball is larger than the force exerted by the surface on the sandbag.

From Newton’s Third Law (that if an object A exerts a force on object B, then object B exerts an equal and opposite force on object A), we can also conclude that the rubber exerts a larger downward force on the surface. This implies that, if the ball hit (say) your hand, then it would hurt more than the sandbag.

Considering change of momentum problems like this helps students answer questions such as the one shown below:

Exam question on change in momentum (solid black arrow and red arrow added)

We can discard options C and D since the change of momentum shown is in the wrong direction: the vertical component of momentum will remain unchanged.

A and B show changes of momentum of the same magnitude in the horizontal direction. However, if we take the horizontal component of the initial momentum as positive then the change of momentum on the gas particle must be negative; this implies that the correct answer is B.

Note also that diagram B shows the pf + (-pi) operation outlined above, with the arrow showing minus pi shown in red (added to the original exam question).