Question

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.

Answer 1

Answer:

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 = $\frac{m1 \times v1 +m2 \times v2}{v(o) - v2}$   ..............2

m = $\frac{6 \times 5 + 4.95 \times 9}{1500 - 9}$  

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 = $\frac{m\times v(o)- m1\times v1}{m}$    

v1 = $\frac{0.0500\times 1500 - 61\times 5}{0.05}$  

v1 = 900 ft/s