Newtonian Problems #4 - Still Struggling

[PhysicsNovice]

I have moved onto kinetic energy and momentum which I thought I understand. Apparently I need more assistance. Please review my questions and answers. Any suggestions and help will be appreciated. Thanks.

1. A sports car with a mass of 1000 kg travels down the road with a speed of 2 m/s. Which is larger, its momentum or its kinetic energy?
a. momentum
b. kinetic energy
c. they are equal
d. the two cannot be compared*

2. Which of the following objects has the largest kinetic energy? A mass of _____with a speed of _________.
a. 6 kg ...2 m/s
b. 4 kg ...3 m/s
c. 2 kg ...5 m/s
d. 6 kg ...3 m/s*

3. Under what conditions is the kinetic energy conserved in a collision?
a. It is always conserved*
b. When the collision is totally elastic
c. When there is no outside force
d. When there is no friction

4. A 3-kg toy car with a speed of 6 m/s collides head-on with a 2 -kg car traveling in the opposite direction with a speed of 4 m/s. If the cars are locked together after the collision with a speed of 2 m/s, how much kinetic energy is lost?
a. 10 J
b. 28 J*
c. 60 J
d. 70 J

5. A tennis ball on the end of a string travels in a horizontal circle at a constant speed. The circle has a radius of 2 m, the ball has speed of 3 m/s, and the centripetal force in 1.5 N. How much work is done on the ball each time it goes around?
a. zero
b. 6 J
c. 10 J*
d. 12 J

6. Which of the following properties of a ball is conserved as it falls freely in a vacuum?
a. kinetic energy*
b. gravitational potential energy
c. momentum
d. mechanical energy

7. If a 0.5-kg ball is dropped from a height of 6 m, what is its kinetic energy when it hits the ground?
a. 3 J*
b. 9 J
c. 30 J
d. There is not enough information to say

8. An elephant, an ant, and a professor jump from a lecture table. Assuming no frictional losses, which of the following could be said just before they hit the floor?
a. They all have the same kinetic energy
b. They all started with the same potential energy
c. They will all experience the same force on stopping
d. They all have the same speed*

9. A 1-kg ball falling freely through a distance of one meter loses 10 J of gravitational energy. How much does the kinetic energy of the ball change if this occurs in vacuum?
a. gain of 1 J
b. gain of 10 J*
c. loss of 1 J
d. loss of 10 J

10. A block of wood loses 160 J of gravitational potential energy as it slides down a ramp. If it has 90 J of kinetic energy at the bottom of the ramp, we can conclude that:
a. mechanical energy is conserved
b. momentum is conserved
c. 250 J of energy was lost
d. 70 J of energy was lost*


PS - Doc Al. I posted this in the college level home forum for four days and only had one viewer. I hope it is ok to post here. I have had much better reponse in doing sol.

 

[NateTG]

1. Correct
2. Incorrect (what's the KE of each?)
3. Incorrect
4. Incorrect
5. Incorrect
6. Incorrect
7. Incorrect
8. Correct
9. Correct
10. Correct
...

The formula for the kinetic energy of an object is:
\frac{1}{2} mv^2

 

[7C0A0A5]

#2) Correct
K=\frac{1}{2} mv^2
(a) K=\frac{1}{2} (6kg)(2\frac{m}{s})^2=12J
(b) K=\frac{1}{2} (4kg)(3\frac{m}{s})^2=18J
(c) K=\frac{1}{2} (2kg)(5\frac{m}{s})^2=25J
(d) K=\frac{1}{2} (6kg)(2\frac{m}{s})^2=27J
Therefore...(d) is the correct answer [if I did it correctly ]

#3) Don't quite me on this one...but I do believe this is incorrect
The answer you are looking for is (b) because in a completely elastic collision the kinetic energies are conversed...in other types of collisions Kinetic Energy is not conserved.

#4) Again don't quote me on this...but...I think you got this one wrong also...I think what you do is find the total Kinetic energy before the collision...then find the kinetic energy after the collision...and subtract the two. Which gives the answer (c)

#6) incorrect [I'm pretty sure I'm right on this one ]
In a "closed system"...kinetic energy is conserved...gravitational potential energy is conserved...Momentum is conserved...
All of these are Mechanical Energies...and Energy is conserved. So the only answer that is totally correct is (d) [note: if I'm wrong, which hopefully I'm not , please correct my mistakes]

#7) not sure again...but I do believe it is incorrect
You want to find Kinetic Energy...But they give you the mass and height.
K=\frac{1}{2} mv^2
But you don't know the velocity. Also:
K=\Delta U_g_r_a_v_i_t_a_t_i_o_n_a_l (U is the Gravitational Potential Energy)
So, find the potential energy at that height...and that should yeild Kinetic Energy. My answer was approximately 30 which is (c)

[If any of this is wrong please correct my mistakes...and I'm sorry if I mislead you in any way. Well hope this helps ]

 

[NateTG]

#2) Correct
(d) K=\frac{1}{2} (6kg)(2\frac{m}{s})^2=27J
Therefore...(d) is the correct answer [if I did it correctly ]

K=\frac{1}{2} (6kg)(2\frac{m}{s})^2=3 kg \times 4 \frac{m^2}{s^2} = 12 \frac{kg m^2}{s^2}
Of course, you did copy it incorrectly, and I was wrong, D is the right answer with 27J - it should be:
K=\frac{1}{2} (6kg)(3 \frac{m}{s})^2=3kg \times 9 \frac{m^2}{s^2}=27 \frac{kg m^2}{s^2}