L: Worksheet Linear momentum; center of mass
1. A big, fat asteroid floats happily in space unaware that another asteroid of mass m = 200 kg is heading towards it at 150 m/s. The big asteroid is initially at rest. The collision breaks the big asteroid in two pieces of mass M = 500 kg that come out in the directions indicated in the figure. The incident asteroid bounces back at an angle as shown in the figure, at 50 m/s. v1 M m 150 m/s 2M Before... 50 m/s 30 Find the speeds v1 and v2 of the two pieces of the big asteroid. Conservation of linear momentum in the x and y directions: x : mv0 = Mv2 - mv3 sin 30 y : 0 = Mv1 - mv3 cos 30 From the first equation: v2 = m ( v0 + v3 sin 30 ) M 200 kg = (150 m/s + ( 50 m/s ) sin 30) 500 kg = 70 m/s m v3 cos 30 M 200 kg (50 m/s) cos 30 = 500 kg = 17 m/s (where v0 = 150 m/s and v3= 50 m/s) m After... M
v2
From the second equation v1 =
(Worksheet L) 2. Romeo (77 kg) sits in the left end of a 2.7 m-long boat (80 kg) that floats still at some distance from the shore and plays a romantic song with his guitar to Juliet (55 kg), who sits in the right end. After the serenade, Juliet carefully moves to Romeo's side to plant a kiss on his cheek.
2.7 m The friction between the boat and the water is negligible. a. What is the net external force on the system boat + Romeo + Juliet? Zero. b. Does the center of mass move with respect to the If shore? yes, in which direction (left/right)? No, it does not move. Since the external force is zero, the acceleration of the center of mass is zero.Thus, the velocity of the center of mass is constant (and this velocity was clearly zero at the beginning). c. Does the boat move with respect to the shore? If yes, in which direction (left/right)? The boat and Romeo move to the right. d. Does Juliet move with respect to the shore? If yes, in which direction (left/right)? Juliet moves to the left e. Find how much does the boat move with respect to the shore. The boat + Romeo system moves to the right and Juliet to the left, in such a way that the center of mass of the whole thing does not move. Let us take x = 0 at Romeo's initial position. After Juliet moves, the boat has shifted some distance d to the right. Then, the positions of the different parts of the system are (see figure)
(Worksheet L)
Romeo
CM for all
boat (CM)
Juliet
x
d
Romeo + Juliet
CM for all
boat (CM)
x
Before: Romeo 0 Juliet: L = 2.7 m CM of the boat: L/2 CM of all: xCM = mR 0 + mb L + mJ L 2 mR + mb + mJ
After: Romeo: d Juliet: d CM of the boat: d+L/2 CM of all: xCM L mR d + mb + d + mJ d 2 = mR + mb + mJ
The position of the CM must be the same in both cases: L L + mJ L = mR d + mb + d + mJ d 2 2 mJ L = mR d + mb d + mJ d mb d=L mJ mR + mb + mJ 55 kg 212 kg
= (2.7 m) = 0.70 m