Skip to main content
Ch 01: Units, Physical Quantities & Vectors

Chapter 1, Problem 37

Relativistic Baseball. Calculate the magnitude of the force required to give a 0.145-kg baseball an acceleration a = 1.00 m/s2 in the direction of the baseball's initial velocity when this velocity has a magnitude of (a) 10.0 m/s; (b) 0.900c; (c) 0.990c.

Verified Solution
Video duration:
10m
This video solution was recommended by our tutors as helpful for the problem above.
345
views
Was this helpful?

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A bullet has a mass of 25.2 g. The bullet is already moving and is to be given an acceleration of 2.5 m per second squared in the same direction as its current velocity. Use relativity to determine the force required when the velocity is I 320 m per second. I I 0.800 C and I I I 0.950 C Awesome. So we're given some multiple choice answers. They're all in the same units of N for both parts I I I and I, I, I, so let's read them off to see what our final answer might be. A is 0. 0.1050, point 202. B is 0.6300 point 1750.646 C is 0.630020. 0.292 2. and D is 0.6650 point and 0.213. OK. So first off, we need to recall the equations for relativistic force. So for, for these equations for relativistic force, there's two, there's two different conditions that we need to consider. We need to consider the parallel condition that we need to consider the perpendicular condition. So let's start with parallel. So for the parallel condition for relativistic force, the equation for that is force is equal to gamma cubed multiplied by the mass multiplied by acceleration. And then for perpendicular conditions, its force equals gamma multiplied by mass multiplied by acceleration. Awesome. So for the parallel conditions, this is where V and F are parallel and then for perpendicular, this is when V and F are perpendicular. OK. So now we need to remember and recall the relativistic factor equation or it's also known as the Lorenz Factory equation. And that states that gamma is equal to one divided by the square root of one minus velocity squared divided by the speed of light squared. Awesome. So gamma, in this case is the Lorenz factor or the relativistic factor. I'm gonna refer to it as relativistic factor equation from now on. OK. So now we could start solving for I. So first off, we need to note that 320 m per second is much less than the speed of light. So we, so thus the relativistic effects are not significant for this equation. So we need to use the relativistic factor equation and the rela relativistic force equation to solve for gamma and to solve for the force F. OK. So let's start by using the relativistic factory equation. Let's plug in all of our known variables and solve for gamma. So one divided by the square root of one minus and then the velocity that were given was 320 m per second. And that's all squared. So 320 m per second squared divided by the speed of light, which the numerical value for that is 3. multiplied by 10 to the eighth power meters per second. And that is squared. OK. So gamma equals one divided by the square root of one minus m per second squared divided by 3.0, multiplied by 10 to the eight power meters per second squared. So when you plug that into a calculator and you should get one, we also can note that 320 m per second squared divided by the speed of light squared is a, is a number that approaches zero. OK. So now we could solve for the force. So let's plug in our known variables. So we know using the perpendicular condition that gamma was equal to one. And now we need to multiply it by the mass. But as we are noticing mass is given to us in grams, but it needs to be in kilograms. So we need to do a quick conversion using dimensional analysis. So 25.2 g multiplied by one kg divided by grams. So that does the conversion and then we need to multiply it by the acceleration which they gave us the acceleration in the problem. And it said it was 2.5 m per second squared. So when you plug that into your calculator, you should get 0. newtons. Awesome. So that is our answer for I. So let's start solving for I I now. So to solve for I I, we need to you like before we need to use the relativistic factor and the relativistic force equation. So let's plug in our known variables solve for the relativistic factor equation. So just like before one divided by the square root of one minus. In this case, they want us to use for the velocity 0.800 C that is squared. So 0.800 C squared divided by the speed of light, which in this case, we're not gonna write out a numerical value. We're just gonna I did AC squared. So we can get the CS to cancel out. OK. So the seas cancel out. And when you plug that into a calculator and if or you don't have to cancel out the CS, you can multiply the 3.0 multiplied by 10 of the power if you would like to. But when you plug that into a calculator, you should get 1. for gamma. So now it's solved for the force like of before. In this case, we need to use the parallel conditions instead of the perpendicular conditions. So we take our gamma value that we just determined which was 1.6667 to the power of three cubed. So did the power three multiplied by than that like. And then we can take, since we know we determine the mass multiplied by the acceleration above we, so we don't have to go through all the work. Again, we could just use our Newton's value that we found above since that's the mass multiplied by the acceleration. So 0. newtons. So when you plug that into a calculator, you should get 0.292 newtons. So that is our answer for I I. So finally, let's start solving for I I I, OK, I I, I, OK. So just like we've been doing before, let's use the same steps we've been using for the past two. Let's use the Lourens factor. So one divided by the square root of one minus. In this case, they gave us the velocity as 0.950 C. So 0. C and that's squared divided by the speed of light squared like of above the seas cancel. And when you plug that into a calculator, you should get 3.2026 per gamma. And then when we solve for f the force, we take our gamma value. And this is also in parallel conditions. So we need to cube it. So 3. cute multiplied by the mass times acceleration which was the value we determined for that was 0.630 newtons. So when you plug that into a calculator, you should get for your final answer. 2.7 newtons. Awesome. And that's our answer for I I I Awesome. Awesome. So that means our final answer must be C I is 0.630 newtons. I I is 0.292 newtons and I I I is 2.7 newtons. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.
Related Practice
Textbook Question
The positive muon (µ+), an unstable particle, lives on average 2.20 * 10^-6 s (measured in its own frame of reference) before decaying. (a) If such a particle is moving, with respect to the laboratory, with a speed of 0.900c, what average lifetime is measured in the laboratory?
486
views
Textbook Question
The positive muon (µ+), an unstable particle, lives on average 2.20 * 10^-6 s (measured in its own frame of reference) before decaying. (b) What average distance, measured in the laboratory, does the particle move before decaying?
392
views
Textbook Question
An alien spacecraft is flying overhead at a great distance as you stand in your backyard. You see its searchlight blink on for 0.150 s. The first officer on the spacecraft measures that the searchlight is on for 12.0 ms. (a) Which of these two measured times is the proper time? (b) What is the speed of the spacecraft relative to the earth, expressed as a fraction of the speed of light c?
446
views
Textbook Question
A proton (rest mass 1.67 * 10-27 kg) has total energy that is 4.00 times its rest energy. What are (a) the kinetic energy of the proton; (b) the magnitude of the momentum of the proton; (c) the speed of the proton?
1147
views
Textbook Question
Electrons are accelerated through a potential difference of 750 kV, so that their kinetic energy is 7.50 * 105 eV. (a) What is the ratio of the speed v of an electron having this energy to the speed of light, c? (b) What would the speed be if it were computed from the principles of classical mechanics?
625
views
Textbook Question
Compute the kinetic energy of a proton (mass 1.67 * 10-27 kg) using both the nonrelativistic and relativistic expressions, and compute the ratio of the two results (relativistic divided by nonrelativistic) for speeds of (a) 8.00 * 107 m/s and (b) 2.85 * 108 m/s.
469
views