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IrinaVladis [17]
4 years ago
11

A strings. A 50 gram bullet, traveling horzontally, strikes the block and becomes embedded inside the block. Immediately after t

he bullet becomes embedded in the block, the block is observed to have a speed of 8.0 m/s. What was the speed of the bullet before it hit the block?
Physics
1 answer:
vazorg [7]4 years ago
5 0

Answer:

u= 200 m/s

Explanation:

Given that

Mass of bullet ,m= 50 gm

Assume that mass of block ,M= 1.2 kg

Lets take speed of the bullet before collision = u m/s

The speed of the system after collision ,v= 8 m/s

There is no any external force ,that is why linear momentum of the system will be conserve.

Linear momentum ,P = mass x velocity

m u = (M+m)v

0.05 x u = (1.2 + 0.05 ) x 8

u= 200 m/s

Therefore the speed of the bullet just before the collision is 200 m/s.

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Answer:

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Explanation:

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3 years ago
An object whose weight is 100lbf( pound force) experiences a decrease i kinetic energy of 500ft-lb, and an increase in potential
alexandr1967 [171]

Answer:

a) 35.75 ft/s

b) 45 ft

Explanation:

<u>Given  </u>

Weight W = 100 lbf

mass(m) = 100*32.174/32.2=99.92 lb

decrease in kinetic energy ΔKE = -500 ft.lbf

increase in kinetic energy ΔPE= 1500 ft.lbf

initial velocity V_1  = 40 ft/s

initial height h_1 = 30 ft/s

The gravitational acceleration g = 32.2 ft/s2 Required  

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<u>Solution </u>

Change in kinetic energy is defined by

ΔKE = .5*m *( V_2 ^2-V_1^2)

Change in potential energy is defined by

ΔKE = W *( h_2 -h_1 )

Then,

-500=.5*99.92*1/32.174*(V_2 ^2-40^2)

V_2=35.75 ft/s

1500 = 100 x (h_2  — 30)  

h_2= 45 ft

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