A 1400 kg blue convertible is traveling south, and a 2080kg red SUV is traveling west. The total momentum of the system consisting of the two cars is 8500 kg.m/s directed at an angle 63.0 west of south.
2007-10-28 07:32:00 · 3 answers · asked by Natiphy2007 1 in Science & Mathematics ➔ Physics
The total momentum = 8500 kg.m/s directed at 63 degrees west of south
The x-component of the momentum = M*sine 63= 7573.551kg.m/s due west
The y component of the same momentum = m*cosin63=3858.91kgm/s due south
For the blue convertible, the momentum = mv due south. This is equal:
1400*v = 3858.91
V = 3858.91/1400= 2.75 m/s
For the red SUV, mv =2080*v= 7573.55. v=7573.55/2080=3.64 m/s
V = 3.64 m/s
2007-10-28 08:14:54 · answer #1 · answered by lonelyspirit 5 · 0⤊ 0⤋
P = (m + M)W; where W is the velocity vector for total momentum P = 8500 kg m/sec. So the magnitude of W is P/(m + M) = W = 8,500/(1400 + 2080) and its direction is theta = 63 deg west of south.
As W = v + V in vector notation, v = W cos(theta) for the south bound car and V = W sin(theta) for the west bound one. Plug and chug.
2007-10-28 08:01:44 · answer #2 · answered by oldprof 7 · 0⤊ 0⤋
tan 63=1400v1 /2080 v2
tan 63 =1.9626=1400v1 /2080 v2
v2=( 0.67344 )v1
[1400v1]^2 + [2080v2]^2=8500^2
v1 = sq rt 18.431=4.293 m/s
speed of blue convertible is 4.293 m/s
v2=2.891 m/s
velocity of red SUV is 2.891 m/s
2007-10-29 03:00:39 · answer #3 · answered by ukmudgal 6 · 0⤊ 0⤋