Conservation of Momentum in 2 Dimensions A 1200 kg car moving at 15 m/s collided
ID: 1408662 • Letter: C
Question
Conservation of Momentum in 2 Dimensions A 1200 kg car moving at 15 m/s collided with a 2500 kg Hummer moving east at 10 m/s. The car bounced off the hummer at 8.0 m/s 10 degree E of S Find the direction of travel of the Hummer immediately of terimpaet A 150 kg pumpkin exploded into three pareses A 60 kg piece fiew due North at 20.0 m/s a 50 kg piece whistled past an un suspecting youth at 30.0 m/s at 30 degree S of E and a third piece struck a stow moving and by stander right in the side of the use a graphical vector approach to predict the direction from the original pumpkin position He by stander was standing when struck by the pumpkin shard. Determine the speed of the pumpkin shard when it impacted the A 6.0 times 10^3 kg is tracking at 300 m/s To after it direction 60. gas is at right angles with speed of 3.60 times 10^3 m/s what is the new velocity of the space chart. Find the final velocity when two air table pucks nit and stick together the first 2.0 kg pucExplanation / Answer
a) initial momentum = final momentum
1200*15 North +2500*10 East = 1200*8( - cos 10 degree North + sin 10 degree East) + 2500*Ve East + 2500Vn North
So, 1200*15 = -1200*8 cos 10 degree + 2500 Vn
Vn = [1200*15 + 1200*8*cos 10 degree ]/2500
= 10.982 m/s
also, 2500*10 =1200*8 sin 10 degree + 2500Ve
Ve = [2500*10 - 1200*8 sin 10 degree]/2500
=9.333 m/s
theta = arctan(9.333/10.982)
= 40.36 degree E of North
part b is not written clearly, especially from 4th line
c) The gas will cause a momentum of 60*3600 kgm/s for the spacecraft in opposite direction
Final momentum of spacecraft = 6000*300 kgm/s i + 60*3600 kgm/s j
velocity = [ 6000*300 kgm/s i + 60*3600 kgm/s j]/6000 = 300 i + 36 j
new speed = sqrt(300^2+36^2) = 302.152 m/s
d) mv = m1v1 +m2v2
v = (2*2.1+2*0.9)/4 = 1.5 m/s
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