Help me out here... 3. An artillery shell is moving on a parabolic trajectory wh
ID: 2162879 • Letter: H
Question
Help me out here...3. An artillery shell is moving on a parabolic trajectory when it explodes in midair. The shell shatters into a very large number of fragments. Choose the correct statement.
a. The force of the explosion will increase the momentum of the system of fragments, and so the momentum of the shell is not conserved during the explosion.
b. The force of the explosion will decrease the momentum of the system of fragments, and so the momentum of the shell is not conserved during the explosion.
c. The center of mass of the system will continue to move on the initial parabolic trajectory until the first fragment touches the ground.
d. The center of mass of the system of fragments will continue to move on the initial parabolic trajectory until the last fragment touches the ground.
e. The center of mass of the system of fragments will have a trajectory that depends on the number of fragments and their velocities right after the explosion.
6. 0.068-kg bullet traveling horizontally at 150.6 m/s strikes a 2-kg block of wood sitting at the edge of a nearly frictionless table. The bullet is lodged into the wood. If the table height is 0.80 m, how far from the bottom of the table does the block hit the floor?
a. 1.2 m
b. 2.5 m
c. 4.9 m
d. 0.5 m
e. 2.0 m
10. A particle is confined by a potential U(x)= 1/x^2 + x - 1, where U is in joules and x is in meters. Determine the net force on the particle when it at x= -1.4 m.
a. 1.7 N
b. -1.9 N
c. -0.3 N
d. 0.3 N
e. -1.7 N
Please provide me a brief explanation for each problem as well.
Explanation / Answer
3. c. The center of mass of the system will continue to move on the initial parabolic trajectory until the first fragment touches the ground
6.conserving momentum :
0.068*150.6=2v
v=10.24m/s
t=sqrt(0.8*2/10) =0.4s
x=10.24m/s*0.4s
=4.096m
10. F=-du/dx
F=2x^-3 -1
F at x= -1.4 m.
= -1.728 N
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