physics help need done soon please ?

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An m = 8 g bullet is fired into a M = 319 g block that is initially at rest at the edge of a frictionless table of height h = 1.1 m. The bullet remains in the block, and after impact the block lands d = 1.7 m from the bottom of the table. Find the initial speed of the bullet. The acceleration due to gravity g is 9.8 m/s ^2 
landing time t = √ 2h/g = √ 2.2/9.8 = 0.474 sec 
block speed Vb = d/t = 1.7/0.474 = 3.59 m/sec 
conservation of momentum shall apply :
bullet speed V = Vb*(M+m) / m = 3.59*327/8 = 147 m/sec 


A m = 4.15 g bullet is fired vertically into a M = 1.98 kg block of wood. The bullet gets stuck in the block, and the impact lifts the block h = 0.074 m up. (i.e. the block  M with the bullet m stuck in it  rises 0.074 m up above its initial position, and then falls back down). Given g = 9.8 m/s ^2 , which was the initial velocity V of the bullet? 
h = 0.074 = Vb^2/2g 
block initial speed Vb = √ 0.074*19.6 = 1.204 m/sec 
conservation of momentum shall apply :
bullet speed V = Vb*(M+m) / m = 1.204*(1984/4.15)  = 576 m/sec 


A rocket is fired vertically upward. At the instant it reaches an altitude of 2720 m and a speed of 296 m/s, it explodes into three equal fragments. One fragment continues to move upward with a speed of 250 m/s following the explosion. The second fragment has a speed of 476 m/s and is moving east right after the explosion. 
What is the magnitude V''' of the velocity of the third fragment in m/s. ?
Suppose a unity mass M worth 1 kg
M = 1*296 North 
M' = 1/3*250  North
M'' = 1/3*476 = 158.7 East 
M''' = 158.7 West + (296-250/3) North 
since M = 1, then velocities are numerically equal to the momentums 
V''' = 158.7 W + 212.7 N 

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