Question

Two astronauts, Hans and Dexter, are on a spacewalk, floating at rest next to their space vehicle. They argue and Hans shoots Dexter with a handgun. The bullet has a mass of 15 grams and a muzzle velocity of 250 m/s. Hans now finds himself drifitng away from the spacecraft.

a. Hans and his spacecraft have a mass of 200kg, how fast is he drifting away?

b. Hans only had one bullet, but he is holding a gun of mass 1.1 kg. How fast must he throw it to stop drifting away from the spacecraft?

c. Meanwhile, on Earth, a solid cylinder of mass 5 kg rolls without slipping toward a hll with a velocity of 5 m/s. How high up the hill will the cylinder roll?

d. The bullet lodges in Dexter, who can be thought of as a long thin rod of length 2 m and mass 200kg. It lodges .8m above his center of mass. What are his linear and angular velocity as he drifts away into space?

Answer 1

part a: as there are no external force, total momentum will remain conserved.

initial momentum=0

hence total final momentum=total initial momentum=0

==>mass of bullet*speed of bullet+mass of hans*speed of hans=0

as bullet's direction and hans' direction is opposite, speed of bullet is taken positive and speed of hans is taken as negative.

==>0.015*250-200*speed of hans=0

==>speed of hans=0.015*250/200=0.01875 m/s

part b:

he has to impart his total momentum unto the gun

hence speed of the gun=mass of hans*speed of hans/mass of gun=200*0.01875/1.1=3.4091 m/s

part c:

let potential energy at ground is 0

let the cylinder rolls till a height of h.

initial mechanical energy=initial linear kinetic energy+initial rotational kinetic energy (as potential energy is 0)

=0.5*mass*speed^2+0.5*moment of inertia*angular velocity^2

moment of inertia of solid cylinder=0.5*mass*radius^2=0.5*5*r^2

total mechanical energy=0.5*5*5^2+0.5*0.5*5*r^2*v^2/r^2

=0.5*5*25+0.5*0.5*5*5^2=93.75 J

at height h, speed will be zero

hence kinetic energy=0

total mechanical energy=potential energy

as there are no energy losses, energy will be conserved.

hence m*g*h=93.75 J

==>h=93.75/(5*9.8)=1.91326 m

part d:

initial linear moemntum of the bullet=0.015*250=3.75 kg.m/s

after being lodged in dexter, total mass of the system=200+0.015=200.015 kg

as momentum is conserved, final linear velocity=initial linear momentum/total mass

=3.75/200.015=0.018748 m/s

angular velocity=linear veloicty/radius

=0.018748/0.8=0.023435 rad/sec