Assuming the second mass sticks with the first mass as it pushes it.
2006-09-27 19:48:11 · 5 answers · asked by Gum Yu Lo 1 in Science & Mathematics ➔ Physics
IF OYU ARE REFERING To A PLASTIC COLLISION THEN
due to conservation of momentum you have
M1*u1 + m2*u2 = (m1+m2)Ucommon
u2 = 0 then
15 *1 = (42 + 15) * Ucommon =>
Ucommon = 15 / 57 = 0.263m/s
2006-09-27 20:13:10 · answer #1 · answered by Emmanuel P 3 · 0⤊ 0⤋
I think the 42 kg mass is too much weight to be pushed by 15kg, then a speed is unlikely, am I right ?
2006-09-28 03:07:11 · answer #2 · answered by Timmo 1 · 0⤊ 0⤋
The final velocity should be 0.51 m/s. This problem may be solved equating the kinetic ebergies before and after the collison.
Kinetic energy is given by 1/2 * m * v^2, where m = mass and v = velocity.
2006-09-28 03:22:52 · answer #3 · answered by bumba_souvik 2 · 0⤊ 1⤋
I think that you should mention with what force are they being pushed. Because otherwise you have thier speed of 1 m/s.
2006-09-28 02:57:21 · answer #4 · answered by nayanmange 4 · 0⤊ 0⤋
They travel at 1 m/s.
Doug
2006-09-28 03:49:54 · answer #5 · answered by doug_donaghue 7 · 0⤊ 0⤋