Question

A 50 grams bullet is fired into a 10kg block that is suspended by a long cord so that it can swing as a pendulum. If the block is displaced so that its center of gravity rises by 10cm, what was the speed of the bullet​

Answer 1

Answer:

Using The Conservation of Energy where

KE= PE or

1/2m(v)^2=mgh

where

m= the combined mass of bullet and wooden block convert 5g mass of the bullet to kg (1000g=1kg) so 5g*1kg/1000g=0.005kg the combined mass= 0.005kg+10kg=10.005kg

v= final velocity of bullet and the wooden block= unknown

g= acceleration due to gravity=9.80m/s^2

h= increase height of the wooden block after the collision with the bullet = 10cm which must be converted to meters (100cm=1m) 10cm*1m/100cm= 0.1m

1/2(10.005)*(v)^2= 10.005*(9.80)*0.1

the 10.005 cancels out of the equation leaving

1/2(v)^2=(9.80)*0.1

multiplying both sides of the equation by 2 gives

(v)^2= 2*(9.80)*(0.1)=1.96

square rooting both sides give

v= sqrt of 1.96

v= 1.4m/s= velocity= speed of mass and wooden block

Using The Conservation Of Momentum

m1v*1(bullet) + m1v1(wooden block)= m(bullet)+m(wooden)*1.4

(the initial momrntum of the wooden =0 kg.m/s)

0.05*v(bullet) +0= 10.005*1.4

v= 10.005*1.4/0.05= 2801.4m/s= speed of the bullet

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