Simple Nature - Light and Matter

as the momentum of the firework before the explosion. However, there is no law of conservation
of kinetic energy, only a law of conservation of energy. The chemical energy in the gunpowder
is converted into heat and kinetic energy when it explodes. All we can say about the kinetic
energy of the pieces is that their total is greater than the kinetic energy before the explosion.
Page 230, problem 45:
Letmbe the mass of the little puck andM= 2.3mbe the mass of the big one. All we need
to do is find the direction of the total momentum vector before the collision, because the total
momentum vector is the same after the collision. Given the two components of the momentum
vectorpx=Mvandpy=mv, the direction of the vector is tan−^1 (py/px) = 23◦counterclockwise
from the big puck’s original direction of motion.
Page 233, problem 62:
We want to find out about the velocity vectorvBGof the bullet relative to the ground, so we
need to add Annie’s velocity relative to the groundvAGto the bullet’s velocity vectorvBA
relative to her. Letting the positivexaxis be east andynorth, we have

vBA,x= (140 mi/hr) cos 45◦

= 100 mi/hr

vBA,y= (140 mi/hr) sin 45◦

= 100 mi/hr

and

vAG,x= 0

vAG,y= 30 mi/hr.

The bullet’s velocity relative to the ground therefore has components

vBG,x= 100 mi/hr

and

vBG,y= 130 mi/hr.

Its speed on impact with the animal is the magnitude of this vector

|vBG|=

√

(100 mi/hr)^2 + (130 mi/hr)^2

= 160 mi/hr

(rounded off to two significant figures).
Page 233, problem 63:
Since its velocity vector is constant, it has zero acceleration, and the sum of the force vectors
acting on it must be zero. There are three forces acting on the plane: thrust, lift, and gravity.
We are given the first two, and if we can find the third we can infer the plane’s mass. The sum
of theycomponents of the forces is zero, so
0 =Fthrust,y+Flift,y+Fg,y
=|Fthrust|sinθ+|Flift|cosθ−mg.