Physics - paper 3

[shorty21]

Could someone please help me with this question :
http://www.xtremepapers.me/CIE/Cambridg ... _qp_32.pdf

Question 1c) part 2 (ii)

thank you

 

[HorsePower]

first, you calculate the increase in P.E. by the formula, (mgh)
the answer would be = 0.6x10x4.0 = 24 J
so before the ball hits the ground all P.E. has converted to K.E. and on impact 22% of it lost.
22% of 24 J is lost. therefore decrease 24 by 22%.
(100-22)/100 x 24 = 18.72 J
the ball has 21.12 J at rebound.
now, again use the formula P.E.= mgh ( where P.E.= 18.72 J , m= 0.6 kg and g= 10 , 'h' is unknown )
18.72 = 0.6 x 10 x h
18.72 = 6 x h
h = 18.72 / 6 = 3.12 m

hope that helps ...

 

[shorty21]

Thank you sooo much

 

[HorsePower]

any time

 

[unplugged]

first, you calculate the increase in P.E. by the formula, (mgh)
the answer would be = 0.6x10x4.0 = 24 J
so before the ball hits the ground all P.E. has converted to K.E. and on impact 22% of it lost.
22% of 24 J is lost. therefore decrease 24 by 22%.
(100-22)/100 x 24 = 18.72 J
the ball has 21.12 J at rebound.
now, again use the formula P.E.= mgh ( where P.E.= 18.72 J , m= 0.6 kg and g= 10 , 'h' is unknown )
18.72 = 0.6 x 10 x h
18.72 = 6 x h
h = 18.72 / 6 = 3.12 m

hope that helps ...

haha , that's quite needlessly long. The energy of the ball when it left the ground is the same as it when hit the ground next time mate ;p.
You just multiply 4 x 0.78 and ta da x).