Linear Momentum — Question 6

Question 6

A tangent to momentum is not a tangent to the path

A particle of mass $2.0\,\mathrm{kg}$ has $\vec p(t)=(2t,4-t^2)\,\mathrm{kg\,m/s}$, with numerical $t$ in seconds.

1. At $t=1.0\,\mathrm s$, find the force and decide whether the speed is increasing.

2. Find the positive time at which the force is perpendicular to velocity.

Question 6 - Solution

A tangent to momentum is not a tangent to the path

(a) Derivation and calculation

Differentiate the given vector history: $$\vec F(t)=\frac{d\vec p}{dt}=(2,-2t)\ \mathrm N.$$ $$\boxed{\vec F(1)=(2,-2)\ \mathrm N},\qquad\vec v(1)=\frac{(2,3)}2=(1,1.5)\ \mathrm{m/s}.$$ $$\vec F\cdot\vec v=2(1)-2(1.5)=-1\ \mathrm W<0,$$ so the speed decreases.

(b) Result and interpretation

Perpendicularity requires $\vec F\cdot\vec p=0$: $$2(2t)-2t(4-t^2)=2t(t^2-2)=0.$$ $$\boxed{t=\sqrt2\ \mathrm s}\quad\text{for }t>0.$$ The nonzero force then changes direction without an instantaneous first-order speed change.

Physical check

The zero power at this time agrees with an extremum of the squared momentum magnitude.