Recent content by aimen khattakk

that is the Pab not the velc OH so since I am looking for the magnitude i don't need to consider their directions so it would just be c2 = 630002+ 507002 c2 = √630002 + 507002 so c = 80867.11 kg⋅km/h ?? so then velocity would be 80867.11 kg⋅km/h / 2700kg so v = 29.9 km/h! thank you! that...

and I am basically looking for the sum of 2 vectors so it would be that the cars are going S of E not just S or just E so doing what i did earlier would have made the direction be just south right?

oh wait, so since I am looking for the overall momentum of Pab after the collision, i don't need to consider negative and positives? since I am not going in a specific direction, rather I am going in 2 directions not one?

my calculator is fine, I've used it for all my courses and I am in 2 other math courses so its not broken, but i took ur advice and tried it with another one and still getting the same answer

i do not understand hot i it is? i don't understand how because i did 507002 + 630002 and since 63000 is actually in the negative direction i basically did 507002 - 630002 and then square rooted it. do i not assign that negative direction? but its going south?

okay so PABY' = 63000kg⋅km/h and PABX = 50700 kg⋅km/h [E] so c2 = a2 + b2 so c2 = 630002 + 507002 c2 = √630002 + 507002 so c = 37396.6576 kg⋅km/h so this is PAB ' now i need to find the velocity after collison so i took the combined mass and divided it with 37396.6576kg⋅km/h and my v...

Homework Statement One car of mass 1400 kg is traveling at 45km/h when it collides and becomes entangles with another car mass of mass 1300 kg traveling at 39km/h[E]. What is their velocity after collision So MA = 1400 kg VA = 45kh/h MB = 1300 kg VB = 39 kh/h [E] and After Collision: MAB =...

I just want to ask one more question so i know i understand. The 7.67m/s that we used, is the final velocity when the ball comes in contact with the ground right? not the initial velocity for when it leaves the ground? thank you for all the help!

IT WORKED! thank you so so much for the help! i finally got it and i understand it too! thank you!

the direction of the Vi would be downward right? and the direction of the Vf would be upwards? so that would make it 7.67m/s-(-8.86m/s) which is 16.53m/s and then i plug that into the Fnet equation and solve for Fn?

okay so here is all the data i have that I am going to lay out Ball dropped from 4m and falls to ground vi = 0 and Vf = 8.86 (calculated using [Vf2 = Vi2 + 2ad]) Ball in contact with ground Vi = 8.86m/s and Vf = Vi of third part which works out to be 7.67m/s Ball bounces to height of 3m Vf...

Okay so let me see if i understand. the final velocity that i calculated earlier (8.86m/s) is now the initial velocity for my second part? using THAT initial velocity, i need to find my final velocity for the second? and id find that using [Vf2 = Vi2 + 2ad]? and then use [Fnet.average = MΔV/Δt]...

so the initial speed is the one i found originally? 8.86m/s? and now using that i need to find my final speed once the ball leaves the floor? so would i do Vf2 = Vi2 + 2ad Vf2 = 8.86m/s2 + 2(-.981m/s2)(3.0m) and then solve it for Vf? i did this and got 4.42m/s for my Vf then when i put that...

Okay so for the Vi i'll need to use the height that the ball was initially dropped from to find the Vi the moment it hits the ground? so would i do the following: d=Vit+2at2 and sub in all the values and then do the same for Vf but use a d of 3.0 instead? because that's how high it bounced...

the 8.86m/s is the final velocity that was calculated in the previous question in order to find the momentum of the ball just before it hits the ground. Now i need to find the normal force acting on the ball after it bounces to a height of 3.0m. so once the ball hits the ground isn't my initial...