I'm 16 year old physics enthusiast , and I wanted to know how to imagine the forces here cuz I was getting stuck there and you cleared it , I'm really good at mathematics so I was able to do the math by myself thanks
*In a ballistic test, a 25g bullet travelling horizontally at 1200m/s pass through a 30cm thick, 350kg stationary target and emerges with a speed of 900m/s. The target slides without friction on a smooth horizontal surface.* I) Calculate the average force(Favg), that the bullet exerts on the target. II) Calculate the target's speed after the bullet emerges.
Thank you for the explanation. I have a doubt regarding the Normal Force component. In inclined plane problems we use mgcos(ß)=N but in uniform circular motion, we use Ncos(ß)=mg. Why is it so? Can you please give a physical explanation for the concept?
Brother because mg acts exactly downwards where as N depends on surface and body contact in Newton laws of motion we represent the equation f=MN where n=mg in order to make n=mg we have to make sure that both are acting opposite to each other but when there is case of curved road N slightly changes so in order to make n=mg we need to resolve the mg in 2components there you can find which component of mg is opposite to n I.e we write n=mgcostheta
Because of the choice of coordinate system. Usually, we choose the coordinate system as parallel and perpendicular to the inclined plane, however, as Physics ninja mentioned, in this case it's useful to consider a coordinate system in which centripetal acceleration is parallel to the x-axis, makes life easier.
Does the friction force where the rubber meets the road acts radially inward to counter the tangential velocity on the car moving in a circular motion?
in an inclined plane,there is gravitational acceleration towards the inclined plane nah.but here you did not mark that force.can you please explain me.iam stuck in that position.thanks
I guess if we decided to rely on the vector components of m*g (force of gravity [otherwise known as gravitational acceleration but that's besides the point]), then we would have to adjust our coordinate system parallel to the incline, which was what Physics Ninja showed wouldn't work *since our acceleration is horizontal in nature when considering the acceleration of this scenario.* I don't really know if the centripetal acceleration being horizontal is just a natural thing when considering these types of problems, but I think they are. Thank you for asking the question, and feel free to follow up with me.
Its just the eq. of velocity where frictional wear and tear would be zero, as we see we have not taken frictional force into consideration in the question...the overall velocity(including mhu, coeff. of friction) would be v=√( (Rg(tanθ + mhu.s) )/ (1-mhu.s tanθ) )..write it down on a paper if you find any difficulty in reading this🤪
In theory yes, but practically speaking the faster you go the larger the Normal force from the road on the car. Clearly this force can only reach a max ir the road will break!!
Muhammad Asim if you look at where the angle is defined with respect to the Normal force you should see that the ramp angle theta is the same angle between the normal and the y-axis. If this is the case the x-comp of normal is Nsin( theta) and the y-comp is y-cos(theta). Simply use the definition of trig functions.. SOH, CAH, TOA
One of the videos of all time
Facts.
I'm 16 year old physics enthusiast , and I wanted to know how to imagine the forces here cuz I was getting stuck there and you cleared it , I'm really good at mathematics so I was able to do the math by myself thanks
We love you physics ninja!
this video is a godsend
*In a ballistic test, a 25g bullet travelling horizontally at 1200m/s pass through a 30cm thick, 350kg stationary target and emerges with a speed of 900m/s. The target slides without friction on a smooth horizontal surface.*
I) Calculate the average force(Favg), that the bullet exerts on the target.
II) Calculate the target's speed after the bullet emerges.
Need help there, bud? Lol.
Great video
Thank you for the explanation. I have a doubt regarding the Normal Force component. In inclined plane problems we use mgcos(ß)=N but in uniform circular motion, we use Ncos(ß)=mg. Why is it so? Can you please give a physical explanation for the concept?
Great question, I will make a video next Wednesday when I get back from vacation
Brother because mg acts exactly downwards where as N depends on surface and body contact in Newton laws of motion we represent the equation f=MN where n=mg in order to make n=mg we have to make sure that both are acting opposite to each other but when there is case of curved road N slightly changes so in order to make n=mg we need to resolve the mg in 2components there you can find which component of mg is opposite to n I.e we write n=mgcostheta
There is not any physical explanation, but there is a mathematical explanation for that and that is the Pythagorean theorem.
Because of the choice of coordinate system. Usually, we choose the coordinate system as parallel and perpendicular to the inclined plane, however, as Physics ninja mentioned, in this case it's useful to consider a coordinate system in which centripetal acceleration is parallel to the x-axis, makes life easier.
cleared a lot of things for me. thanks!
THANK YOU
Does the friction force where the rubber meets the road acts radially inward to counter the tangential velocity on the car moving in a circular motion?
Friction acts parallel to the road.
much appreciate
i dont get how you can get angle theta in the triangle
Thanks! Very helpful! New subscriber here.
in an inclined plane,there is gravitational acceleration towards the inclined plane nah.but here you did not mark that force.can you please explain me.iam stuck in that position.thanks
I thought you would have to use Pythagoras to find the normal force, I didn't realise you can just add components to get the resultant vector?
why is there no angles for the force of gravity?
I guess if we decided to rely on the vector components of m*g (force of gravity [otherwise known as gravitational acceleration but that's besides the point]), then we would have to adjust our coordinate system parallel to the incline, which was what Physics Ninja showed wouldn't work *since our acceleration is horizontal in nature when considering the acceleration of this scenario.*
I don't really know if the centripetal acceleration being horizontal is just a natural thing when considering these types of problems, but I think they are. Thank you for asking the question, and feel free to follow up with me.
great one thx man
for the forces in the x-direction, why is Nsin not negative?
I alway pick the direction that is toward the center of the circle to be positive for circular motion problems.
is the final answer the minimum speed?
Its just the eq. of velocity where frictional wear and tear would be zero, as we see we have not taken frictional force into consideration in the question...the overall velocity(including mhu, coeff. of friction) would be
v=√( (Rg(tanθ + mhu.s) )/ (1-mhu.s tanθ) )..write it down on a paper if you find any difficulty in reading this🤪
Hi, does that mean that at 90 degrees you can go infinitely fast?
In theory yes, but practically speaking the faster you go the larger the Normal force from the road on the car. Clearly this force can only reach a max ir the road will break!!
@@PhysicsNinja thanks
Why at y co coordinate use cos and on x coordinate use sin? Explain.....
Muhammad Asim if you look at where the angle is defined with respect to the Normal force you should see that the ramp angle theta is the same angle between the normal and the y-axis. If this is the case the x-comp of normal is Nsin( theta) and the y-comp is y-cos(theta). Simply use the definition of trig functions.. SOH, CAH, TOA
👍
clutch
cos