An 8.11g bullet is fired into a 325g block that is initially at rest at the edge

Question

An 8.11g bullet is fired into a 325g block that is initially at rest at the edge of a frictionless table of height h = 1.13 m

a)The bullet remains in the block, and after the impact the block lands d = 2.12 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?

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

c)Determine the initial speed of the bullet.

An 8.11g bullet is fired into a 325g block that is initially at rest at the edge of a frictionless table of height h = 1.13 m a)The bullet remains in the block, and after the impact the block lands d = 2.12 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? b)What is the initial horizontal velocity of the block as it flies off the table? (assume this to be in the positive direction) c)Determine the initial speed of the bullet.

Explanation / Answer

part A: time of fall can be obtained by the kinematic equation

S = ut + 0.5 at^2

where S is distance

a is accleration

t is time of fall

here u = 0

so

S = 0.5 at^2

t^2 = 2S/a

t^2 = 2* 1.13/9.8

t^2 = 0.2306

t = 0.48 secs
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part B:

in 0.48 secs is travels a distance of 2.12 m

so in 1 sec, the block moves a distance of 2.12/0.48   = 4.416 m/s

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apply the law of conservation of momemtum as

MVof bullet = (m+M) V of block

so

Vbullet = (0.325 + 0.00811) * 4.414/(0.00811)

Vbullet = 181.3 m/s

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