Two friends, Al and Jo, have a combined mass of 158 kg. At an ice skating rink they stand close together on?

Question

skates, at rest and facing each other, with a compressed spring between them. The spring is kept from pushing them apart, because they are holding each other. When they release their arms, Al moves off in one direction at a speed of 0.910 m/s, while Jo moves off in the opposite direction at a speed of 0.829 m/s. Assuming that friction is negligible, find Al's mass

Answer 1

let al' mass be m

conserving momentum
m*0.91=(158-m).829
solving for m we get
m=75.320kg

Answer 2

Mv - mV = (M + m)v0 = 0 from the conservation of momentum; so M = mV/v where M = big person mass, m = small person mass, V = small person velocity, v = big person velocity. v0 = the velocity of the M + m two masses while together, which is zero.

Thus, we have M = m(V/v) and we also have M + m = K = 158 kg. So combining, there is m(V/v) + m = K; and m(1 + V/v) = K so that m = K/(1 + V/v), which you can now find because K = 158, V = .910 and v = .829.

Once you find m, put that in M + m = 158 and solve for M. Jo is obviously a little piggy because she clearly weighs more than Al...tsk, tsk.

To points to learn here: the conservation of momentum, the sum of momenta of two or more parts equals the momentum of the parts before splitting off, and the fact that more massive bodies (e.g., Jo) will move slower than the lighter ones (e.g., Al) after they split away.

Answer 3

enable their hundreds be A kg (for Al) and J kg (for Jo). A + J = 168------------------------- (a million) initially at relax, so preliminary momentum = 0 very final momentum = A (0.eighty 5) + J (-a million.4), considering that opposite course. making use of concept of conservation of Momentum, 0.85A - a million.4J = 0 0.85A = a million.4J J = (0.85A / a million.4) ----------------------(2) Sub (2) into (a million), A + (0.85A / a million.4) = 168 appearing some algebra, 2.25A = 235.2 A = 104.5 kg (ans)

Answer 4

let Al 's mass be m
from consertvation of momentum we have
m*.910=[158-m]*.829
m[.910+.829]=158*.829
m
=158*.829/1.739
=75.32kg