1. The cart below (including the cannon and the ball) has a total mass of 100 kg

Question

1. The cart below (including the cannon and the ball) has a total mass of 100 kg and a velocity of 3 m/sec to the right. The cannon ball (mass- 10 kg) is fired at a speed of 30 m/sec to the left. a) What is the final velocity of the cart (whose total mass is now 90 kg)? Initial Final 30 m/sec Vi b) Assuming no heat loses, what is the initial PE stored in the gun powder? 2. Three masses collide and stick together. Their velocities and masses are given below What is the final speed of the three masses stuck together? m -3kg VA -5 m/sec ma 2 kg Vi-10 m/sec mc 4 kg Ve 6 m/sec The angle between B and C is 45 co

Explanation / Answer

1.) Initial momentum = Final momentum

100 x 3 = 90 x vf - 10 x 30

90 vf = 300 + 300 = 600

vf = 6.6666667 m/s

b.) Initial PE stored in the gun powder = Final KE - Initial KE = 0.5 x 90 x vf2  + 0.5 x 10 x 302 - 0.5 x 100x 32

PE = 0.5 [ 90 x 6.6666672 + 10 x 900 - 100 x 9 ] = 6050 Joules

2.) va = -5 j m/s mava = - 15 j kgm/s

vb = 10 i m/s mbvb = 20 i kgm/s

vc = ( 6Cos 45 i + 6Sin 45 j) m/s mcvc = ( 16.97 i + 16.97 j ) kgm/s

if v is the final velocity of the combined mass, according to the law of conservation of linear momentum,

mava + mbvb + mcvc = ( ma +mb + mc ) v

- 15 j + 20 i + ( 16.97 i + 16.97 j ) = 9 v

36.97 i + 1.97 j = 9 v

v = 4.12 i + 0.2188 j m/s

3.) va = 6 i m/s mava = 12 i kgm/s

vb = 0 m/s mbvb = 0 kgm/s

according to the law of conservation of linear momentum,

mava + mbvb = mavaf   + mb vbf

12 i = 2 ( 5 Cos 53 i + 5 Sin53 j ) + 4vbf

6 i - 8 j = 4vbf

vbf = 1.5i - 2j ms

magnitude = 2.5 m/s

angle = tan-1 (-2 /1.5) = - 53 degrees

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