Recent content by cooney88

AW DUDE! couldn't we just use m1xv1+ m2xv2 = m1xv1+m2xv2 ? i think the angle might of been in the question to put u off?

which i did but I am still getting 4.2 . thanks a million for helping me with this by the way. still not sure what I am doing wrong tho?

yes only before the collision are v1 and v2 are different. however the 2 balls ''stick together'' therefore the velocity is the same for both of them after the collision

[b]1. an ice cone 25mm high with a max radius of 15mm is floating apex downwards in a glass of water. if the ice cone has a mass of 5.3g to what depth will the cone be immersed? density of water is 1000kg/m^3 i know from the law of flotation that the weight of a floating body is equal to...

sorry i meant i got 4.2 not 4.02 . so I am obviously doing something wrong

any ideas what I am doing wrong?

pax= 6v1 pay = 0 pbx= 8v2cos30 pby = 8v2sin30 pcx= 6v1 +8v2cos30 = 12.9v pcy = 8v2sin30= 4v (12.9v^2 + 4v^2)^.5 = 13.5v so 58 = 13.5v v= 4.2 a is moving along the x-axis and b is moving below a but is heading towards a. the angle between them is 30

doc al could you take a look at my query please: its doing my head in. https://www.physicsforums.com/showthread.php?t=314056

cheers buddy

so its the kinetic formula cos it wouldn't make sense to use the hookes law right? but since when can u have the spring constant in a kinetic energy formula?

Q. Two objects A and B are moving on a frictionless horizontal surface. A has a mass of 6kg and speed of 3m/s , B has a *** of 8kg and a speed of 5m/s. if they collide and stick together the speed in m/s of the composite object after the collision is: diagram: A----------------------> X...

Q. An object of mass o.4kg slides at 8m/s across a frictionless path before striking one end of a spring that is fixed at the other end. the spring constant is 80N/m. the maximum distance, in metres, by which the spring compressed is : ans? I a not sure wether to use hookes law (f=ks)...