An 8.88g bullet is fired into a 295g block that is initially at rest at the edge

Question

An 8.88g bullet is fired into a 295g block that is initially at rest at the edge of a frictionless table of height h = 1.07 m (see the figure below).

1. The bullet remains in the block, and after the impact the block lands d = 1.85 m from the bottom of the table. How much time does it take the block to reach the ground once it flies off the edge of the table?

2. What is the initial horizontal velocity of the block as it flies off the table? (assume this to be in the positive direction)

3. Determine the initial speed of the bullet.

Explanation / Answer

Mass of bullet = 8.88g

Mass of block = 295g

height = h = 1.07m

We know that

Time of flight = T = 2Vo sin theta / g............a

Angle =theta=?

From the above given data

Range =d = 1.07m

Height = h=1.07m

Range =d= Vo2 sin 2theta / g .............1

Height = h = Vo2 sin2 theta / g.........2

dividing equation 1 by equ2

d / h = sin 2 theta / sin2 theta

d / h = 2sin theta . cos theta / sin2 theta

d / h = 2 cos theta / sin theta

h / d = 2 sin theta / cos theta

h / d = 2 tan theta

theta = tan-1 h / 2d

theta = tan-1 1.07 / 2(1.05)

theta = 26.970

Dealing with equation a , the variable still unknown is horizontal velocity = Vo = ?

In order to find the horizontal velocity , putting the value of theta in equ1

d = V2o sin 2 theta / g

1.85 = Vo2 sin 53.95 / 10

2).. V =4.81m/s ( This is the value of horizontal velocity of composite system)

Putting the vlaue of theta and V in equ a

T = 2(4.81) sin 26.97 / 10

1) T = 0.43sec

3) Initial speed of bullet = Vi = ?

By the coservation of momentum

Mb Vb + MoVo = (Mb + Mo) V

Vb = (Mb + MO) V / Mb

Plugging in the values

Vb = (8.8g + 295g) 4.81 / 8.8g

Vb = 166 m/s

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