Problem Of the week: 20th November 2023
One of the CCF’s guns is designed such that it can launch a projectile of mass 10 kg at a speed
of 200 m/s. The gun is placed close to a straight, horizontal railway line and aligned such that the projectile will land further down the line. A small rail car of mass 200 kg travelling at a speed of 100 m/s passes the gun just as it is fired. Assuming the gun and the car are at the same level, at what angle upwards must the projectile be fired so that it lands in the rail car?
Credit: University of Oxford Physics Problems
Draw a clear diagram to set up the problem. Do you need all the information that the question gives you, or are some pieces of information irrelevant?
For projectile/SUVAT questions such as these it is important to look for information that is hidden in the question. For example, there is a vertical acceleration of a = −g = −9.81 m/s2 implicitly present in the problem, even though it isn’t mentioned. You are also expected to assume there is no air resistance, so there will be no horizontal accelerations of any kind. Likewise, although you don’t have to do this explicitly, it is worth thinking about what the following pieces of information tell us:
“Passes the gun just as it’s fired”: we can set t = 0 at this point and then use the same t coordinate for both the rail car and the projectile.
“Assuming the gun and the car are at the same level”: we don’t have to worry about any vertical displacement or offset between the two objects.
Finally, given that this exists in the world of projectiles, in which objects are in free-fall, the masses of the objects are irrelevant. Their inclusion in the question is merely a red herring.
A clear diagram is often the best way to lay out all of the information given to you in a problem – see Figure 1.1.
Figure 1.1: A diagram showing the gun firing a projectile of mass m at an angle θ to the horizontal and at a speed of 200 m/s at the same time as the rail car of mass M passes by at 100 m/s
There are two points at which the projectile and the car are in the same position:
1. When the projectile is launched (let this be t = 0)
2. When the projectile lands in the car (let this be t = t)
Upon landing the two objects therefore have the same horizontal displacement, sp = ss. It is worth adopting straightforward notation: any quantity with a subscript p refers to the projectile, and any quantity with a subscript c refers to the car.
The horizontal acceleration of the rail car is zero, and the vertical acceleration of the projectile, if we take upward to be positive, is ap = −9.81 m/s2.
Note how we’ve managed to “pull” a lot of information out of the question that we weren’t explicitly given.
Let θ be the angle above the horizontal at which the projectile is launched.
The horizontal speed of the projectile is then up = 200 cos θ m/s and the horizontal speed of the car is uc = 100 m/s. Given that they must both travel the same distance in the same time for one to land in the other:
sp = sc
upt = uct
up = uc
Substituting up = 200 cos θ m/s and uc = 100 m/s and simplifying:
200 cos θ = 100
cos θ =100/200
θ = cos-1(1/2)
θ = 60°
Notice that the way we have solved this allows us to appreciate the general fact that the horizontal velocities of the car and the projectile must be the same no matter what the other parameters of the question. It is worth thinking about why this makes sense: assuming no air resistance there is no horizontal acceleration in the question, so for the two objects to both start and end at the same horizontal displacement they must always be travelling at the same horizontal velocity.