Kaplan Science Assessment Physics Q3

[betterfuture]

---

Members do not see this ad.

A projectile of mass 20 kg is shot at an angle of 60 degrees to the horizontal at a speed of 100 m/s. What is the energy of the projectile at its highest point?

Answer: 1 x 10^5 J

They explain the answer using the KE equation. I am just confused as to how KE is used since I thought speed at the highest point or max height is 0. Thus KE=0. Am I missing something here? Thank you in advance.

 

[aldol16]

They explain the answer using the KE equation. I am just confused as to how KE is used since I thought speed at the highest point or max height is 0. Thus KE=0. Am I missing something here? Thank you in advance.

You're confusing two different scenarios. KE = 0 at highest point only applies for vertical projectiles. Shoot something up and eventually it will have to come back down, meaning the velocity changes signs and thus must go through zero.

In this case, you're shooting something at an angle - that is, with both vertical and horizontal velocities. The vertical component will be zero at the highest point but the horizontal component won't. Otherwise, a cannonball would just fly out of the cannon, reach the high point, and then drop straight to the ground.

 

[theonlytycrane]

A projectile of mass 20 kg is shot at an angle of 60 degrees to the horizontal at a speed of 100 m/s. What is the energy of the projectile at its highest point?

Answer: 1 x 10^5 J

They explain the answer using the KE equation. I am just confused as to how KE is used since I thought speed at the highest point or max height is 0. Thus KE=0. Am I missing something here? Thank you in advance.

Just curious- does the question ask for the kinetic energy at the highest point or total energy?

 

[betterfuture]

Okay so this is their explanation:

At the highest point, the projectile will possess two types of mechanical energy: potential energy, because of its height and kinetic energy, because of its speed. The total mechanical energy of the projectile is equal tot the sum of the kinetic and potential energies. This question also doesn’t indicate any dissipative forces acting, which means that thet total energy at the highest point will be equal to the energy at the beginning when all energy is kinetic. The equation for kinetic energy KE=1/2 mv^2 = ½(20)(100)^2=(10)(10000)=1 x 10^5

You're confusing two different scenarios. KE = 0 at highest point only applies for vertical projectiles. Shoot something up and eventually it will have to come back down, meaning the velocity changes signs and thus must go through zero.

In this case, you're shooting something at an angle - that is, with both vertical and horizontal velocities. The vertical component will be zero at the highest point but the horizontal component won't. Otherwise, a cannonball would just fly out of the cannon, reach the high point, and then drop straight to the ground.

So quick question, what is the angle given for? I thought they would multiply the speed 100 m/s by cos60 degrees to get the horizontal component. I am just confused to why their answer did not take this into account.

 

[betterfuture]

Just curious- does the question ask for the kinetic energy at the highest point or total energy?

Actually, that is their question, what I wrote in my post.

What is the energy of the projectile at its highest point?


I would assume it means the total energy at the highest point

 

[aldol16]

So quick question, what is the angle given for? I thought they would multiply the speed 100 m/s by cos60 degrees to get the horizontal component. I am just confused to why their answer did not take this into account.

The angle is given so that you can calculate energy if you wanted to determine speed at the highest point. Your approach is valid in that energy is conserved. But there's also enough information for someone who didn't see that and instead went the roundabout way of doing it, calculating the maximum height and its associated potential energy and then the kinetic energy at that point as well.

 

[betterfuture]

It's still not clear to me why the answer is what it is.

 

[aldol16]

It's still not clear to me why the answer is what it is.

It's a conservation of energy problem. The total energy of the system is conserved at all points. So one way to approach it is to say the energy is always equal to the initial energy, which was all kinetic and equal to 0.5(20)(100)^2. Another way to do it is to imagine the energy conversions. When the projectile is shot, it has all kinetic energy. Once it reaches the highest point of its flight, some of that energy has been converted to potential energy and the rest remains kinetic energy in the horizontal direction. Once you realize that, you can again apply conservation of energy or you can also solve for the max. height to find the potential energy and add that to the kinetic energy at that point. Either way, you'll get the same answer because you can't ever create or destroy energy.

 

[betterfuture]

Okay, I get that. When the mass is initially projected, it has KE and no PE, and it mentions no dissipative forces, hence Energy is conserved.

Okay. But my question is, when calculating KE of a system, like for this example for instance, the speed used in the equation KE=1/2mv^2, is the speed they give you? And you don't take into account the angle for figuring out the speed? Is that for all types of problems like the one given.

 

[aldol16]

Okay. But my question is, when calculating KE of a system, like for this example for instance, the speed used in the equation KE=1/2mv^2, is the speed they give you? And you don't take into account the angle for figuring out the speed? Is that for all types of problems like the one given.

You can approach this question in one of two ways. You can realize that the "angle" is just an arbitrarily defined direction. In other words, you can just make your x-y axis where your x-axis lies along the angle they shoot the ball at. Energy is still conserved. Or, you could split it up into components if you want. You could say horizontal component is 100*cos*60 and vertical component is 100*sin60. Thus, vertical kinetic energy is 0.5*m*(100*cos60)^2 and horizontal kinetic energy is 0.5*m*(100*sin60)^2. Then, total energy is 0.5*m*((100*cos60)^2+(100*sin60)^2, which is equal to 0.5*m*100^2*(cos^2(60) + sin^2(60)). Then just realize that cos^2 + sin^2 = 1 (trig identity) and your answer reduces to the original.