The impossible physics problem It\'s not 0.01 m. It\'s not 3.3 m. Any attempt to

Question

The impossible physics problem

It's not 0.01 m. It's not 3.3 m. Any attempt to solve it thus far has been incorrect.

A 0.0270 kg bullet moving horizontally at 400 m/s embeds itself into an initially stationary 0.500 kg block. a) What is their velocity just after the collision? 20.49 (b) The bullet-embedded block slides 8.0 m on a horizontal surface with a 0.30 kinetic coefficient of friction. Now what is its velocity? m/s 19.3 m/s (c ) The bullet-embedded block now strikes and sticks to a stationary 2.00 kg block. How far does this combination travel before stopping?

Explanation / Answer

a) Initial momentum of system of system = 0.027*400 = 10.8 kg m/s

Let velocity after embedment be v

Total mass = 0.027+0.5 = 0.527kg

Final momentum of system = 0.527v

0.527v = 10.8

v= 10.8/0.527 = 20.49 m/s

b) coefficient of kinetic frition = 0.3

deceleration experienced by bullet-embedded block = 0.3*9.8 = 2.94 m/s2

u=20.49 m/s

let velocity after travelling 8 m be v

v2 = u2 + 2as

v2 = 20.492 +(2*2.94*8)

v = 19.3 m/s

c)Momentum of system before striking 2 kg block = 0.527*19.3 = 10.17 kg-m/s

Let velocity of system just after collision be u

Mass of system = 0.527+2 = 2.527kg

u=10.17/2.527 = 4.02m/s

Deceleration on system = 0.3*9.8 = 2.94 m/s2

Let distance travelled by system before stopping be s

s = 4.022/(2*2.94) = 2.75 m

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