A 12-g bullet moving horizontally strikes and remains in a 3.0-kg block initiall

Question

A 12-g bullet moving horizontally strikes and remains in a 3.0-kg block initially at rest on the edge of a table. The block, which is initially 80 cm above the floor, strikes the floor a horizontal distance of 120 cm from its initial position. What was the initial speed of the bullet?

a. 0.68 km/s

b. 0.75 km/s

c. 0.81 km/s

d. 0.87 km/s

e. 0.41 km/s

At t = 0, a wheel rotating about a fixed axis at a constant angular acceleration has an angular velocity of 2.0 rad/s. Two seconds later it has turned through 5.0 complete revolutions. What is the angular acceleration of this wheel?

a. 17 rad/s2

b. 14 rad/s2

c. 20 rad/s2

d. 23 rad/s2

e. 13 rad/s2

Explanation / Answer

t = sqrt(2*y/g) = sqrt(2*0.80/9.8) = .40406 sec.

v'x = x/t = 1.20/.40406 = 2.96986 m/sec

and pi = pf

mbvo = (mb+M)v'b

vo = (mb+M)v'b/(mb)

vo = (0.012+3)*2.96986/0.012

vo = 745.435 m/sec --> ans

2nd problem:

5 revolutions = 31.416 radians

31.416/(2/2) = (vi + vf)

31.416 - vi = vf

vf = 31.416 - 2 = 29.416 rad/s

(vf - vi)/t = a

a= (29.416 - 2)/2 = 13.7 rad/s^2

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