Do Trucks with the Same Kinetic Energy Have Different Momenta?

[hahaha]

I have two questions that I need answers for:

1. A small truck and a large truck have the same kinetic energies. Which truck has the greater momentum? Justify your answer.

2. (a) Can an object have kinetic energy, but no momentum? Can an object have momentum, but no kinetic energy? Explain.
(b) Repeat (a) for an isolated system of two interacting objects.


Thanks in advance for your help.

 

[Atomos]

With regards to question 1, I may be wrong, for I have never taken a course specifically relating to physics in my life:

$$KE=1/2mv^2$$
so, they both have a different mass, so their velocity must also be different between the two vehicle. momentum varies directly to KE
$$p=mv$$
$$1/2v(mv)=KE$$
logically, the smaller vehicle is going to have a higher velocity than the larger vehicle to maintain the same KE.
So, if the $$1/2v$$ factor is then larger for the smaller vehicle, and smaller for the larger vehicle, then the momentums (the factor mv) must be smaller for the smaller vehicle, and larger for the larger vehicle.

 

[Atomos]

Oh and for 2 :a:
Again, forgive if I am wrong, but:
For the momentum to be $$0$$, one or both of the factors must be 0
$$0=mv$$
m or v or both must be 0
if either m or v are 0
$$KE=1/2mv^2$$
if any of the factors equate to 0 the entire thing is 0.
for 2 :b: I am not sure what it is asking.

 

[Jameson]

$$K = \frac{1}{2}mv^2$$

Let's say the smaller truck has half of the mass. $1/2m_2=m_1$

Set up the equation and solve for v, the "masses" should cancel.

 

[hahaha]

Jameson, so are you agreeing with Atomos that the larger vehicle will have a greater momentum?

 

[StatusX]

The easiest way to do the first problem is to use KE=p^2/2m. Since these fractions have to be equal for both trucks, the one with the bigger denominator (mass) must have a bigger numerator.

This also helps with 2 a. For 2 b, the KE is always greater than or equal to 0, and only equal to 0 when the velocities are all 0 (which means momentum is 0). See if you can find a case where the total momentum of the system (which can also be thought of as the velocity of the center of mass times the total mass) is 0, but the individual particles are moving.

 

[TsunamiJoe]

on 1 you don't even need the KE truely, using the P= M*V equation you see that the larger mass will give you the larger momentum, so you need not substitute numbers in unless it is neccesary to visualize, also just think instead of 2 trucks of differint masses, just think of a compact car and a semi, though its more extreme, it shows to a greater effect on the reality that the more wieght the more momentum

P=M*V so in otherwords if velocity is zero, then it has no momentum, so for 2 a and b just say that no they cannot have 0 momentum or 0 velocity and have KE

 

[ChemRookie]

Equilibrium

nm, mistake.

 

[Jayboy]

I like StatusXs explanation, but if you want to work with KE, here is the most explcit way to deal with what you have been given.

Call the heavier truck 1, and the lighter truck 2. Since KE is the same

$$\frac{1}{2}m_{(1)}V_{(1)}^{2} = \frac{1}{2}m_{(2)}V_{(2)}^{2}$$

Since $$m_{(1)} > m_{(2)}$$

we must have $$V_{(1)} < V_{(2)}$$

Cancelling the 1/2s and moving one of the V_(1)s gives:

$$m_{(1)}V_{(1)} = m_{(2)}V_{(2)}(\frac{V_{(2)}}{V_{(1)}})$$

Since we have argued that $$V_{(1)} < V_{(2)}$$ the fraction must be greater than one and we must therefore have:

$$m_{(1)}V_{(1)} > m_{(2)}V_{(2)}$$

For the 2nd part:

The answers have been given by others here. This all deals with the difference between scalars and vectors and it's worth developing a good understanding of this.