If mass is added to a toy car (29.7g) and dropped down a wooden ramp would it affect its speed making it go faster? I know friction comes in to play to, so if you could give me an answer or an equation to show this that would be.
[Physics] If mass is added to a toy car does it affect its speed making it faster
frictionhome-experimentnewtonian-gravitynewtonian-mechanics
Related Solutions
As it turns out, adding mass (while keeping the dimensions of the car fixed) will make the car go faster. This sounds contrary to what all physics students are taught, but the reason is while friction scales with mass, air resistance doesn't. That's why a thirty-foot rock will fall faster than a thirty-foot parachute in an atmosphere. It's also why the winning car in pine-wood or soap-box derby contests usually seem to be right at the weight limit.
Other factors that will affect the car's speed include:
- The mass of the wheels. It takes a little bit of energy to get the wheels spinning (rotational kinetic energy), and the less mass they have the less energy they will absorb.
- You want an axle with a very small diameter. The upward force from the wheel to the axle isn't affected by the axle diameter, but the drag torque (essentially friction) exerted by the wheel onto the axle will be less the smaller the diameter of the axle. If the car is large enough, you could go with a ball-bearing wheel/axle bearing, like real cars or roller blades, since rolling friction is less lossy than sliding friction.
- If you can get a spring-like suspension on the wheels, or use air-inflated tires, that might help. Rigid wheels on a rigid suspension will cause the car to bounce up and down whenever it hits a tiny bump, wasting energy. A good suspension will compress when it rises over an obstacle (slowing the vehicle down slightly), but that energy will be recovered when the wheel rolls down the other side of the bump (allowing the vehicle to speed back up).
You've already mentioned making the car as aerodynamic as possible. If it's running down a track where it might bump up against a side guardrail or something, using a low-friction material will help (Teflon, or the kind of plastic used in gallon milk jugs, or even tiny wheels mounted on the sides). That's all I can think of at the moment. Good luck.
For sure the object you will put in cart will act as one body until and unless they both have same velocity or I say that they both move relatively and this is possible if both have contact with each other. And if you want to take internal forces then you have to work on that but besides everything of what you ask the answer is simple that you have to take that as single body.
Example-
Let us say one block sliding down and for the sake of question and imagination let us take one small section of (red portion in image) as chocolate bar. So by seeing the FBD we can recognize that it will act as one body (as the red section is already a part of that block) will work as one system and it will act just same in your Cart example due to which the mass of the system increases.
Best Answer
Yes, adding mass to a toy car should at least in principle make it accelerate down a ramp faster.
The total force on the car is in the "forward" direction, with magnitude $$F=m g \sin\theta\ -\ m g C_{rr}\cos\theta\ - \tfrac12 \rho v^2 C_D A\ ,$$ where $m$ is the car's mass, $g$ is the acceleration due to gravity at Earth's surface, $\theta$ is the angle from horizontal of the ramp, $C_{rr}$ is the rolling resistance coefficient, $\rho$ is the density of air, $v$ is the speed of the car, $C_D$ is the drag coefficient of the car, and $A$ is the car's cross section area. $C_{rr}$ depends on a lot of things rather than being a constant, but what's important here is that for rigid plastic tires, $C_{rr}$ should decrease with increasing $m$. $C_D$ is independent of $m$.
The first term in the above equation is the forward component of the force purely due to gravity, the second term accounts for rolling resistance, and the third term accounts for drag.
If you equate that equation with $F=ma$ and divide both sides by $m$, you get that the car's acceleration in the forward direction is
$$a=g \sin\theta\ -\ g C_{rr}\cos\theta\ - \frac{ \rho v^2 C_D A}{2 m}\ .$$
According to that equation, if $m$ increases, so does $a$.