A bullet is fired with a horizontal velocity of 1500 ft/s through a 6-lb block A and becomes embedded in a 4.95-lb block B. Knowing that blocks A and B start moving with velocities of 5 ft/s and 9 ft/s, respectively, determine (a) the weight of the bullet, (b) its velocity as it travels from block A to block B.

weight of the bullet is 0.0500 lb

velocity as it travels from block A to block B is 900 ft/s

Explanation:

given data

horizontal velocity = 1500 ft/s

mass block A = 6-lb

mass block B = 4.95 lb

blocks A velocity = 5 ft/s

blocks B velocity = 9 ft/s

solution

we apply here law of conservation of momentum that is

m × v(o) + m1 × (0) + m2 × (0) = m × v(2) + m1 × v1 + m2 × v2     ................1

put here value and we get m

m =   ..............2

m =  

m = 0.0500 lb

and

when here bullet is pass through the A block then moment is conserve that is

m × v(o) + m × v1 + m1 × v1   ............3

v1 =    

v1 =  

v1 = 900 ft/s

so easy. just google the electromagnetic waves and it’ll pop up.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west