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CasedIt
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#1
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stuck on momentum q ples help:

A ball of mass 0.050KG rests on some tennis strings and is held horizontally

a i calculate the difference in magnitude between the two forces on the ball when the racket is accelerated upwards at 2ms

( F = ma ? )

b i The ball is dropped from rest at a point 0.80 m above the racket head

ii calculate the speed of the ball before impact

iii calculate the momentum of the ball before impact

iiii change in moment of ball during impact

iv the average force during impact with contact time of 0.050 s
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TheWolf
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(Original post by CasedIt)
stuck on momentum q ples help:

A ball of mass 0.050KG rests on some tennis strings and is held horizontally

a i calculate the difference in magnitude between the two forces on the ball when the racket is accelerated upwards at 2ms

( F = ma ? )

b i The ball is dropped from rest at a point 0.80 m above the racket head

ii calculate the speed of the ball before impact

iii calculate the momentum of the ball before impact

iiii change in moment of ball during impact

iv the average force during impact with contact time of 0.050 s
xTinaA will help you
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DazYa
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ok i will go through each part for you

first part

F = ma

0.050 x 2
= 0.1 N

second part

speed PE = mfg = 1/2 mv2

v2 = 2gh

v = square root 2gh

= 2 x 9.81 x 0.8-
= 3.96

third

momentum is mv

0.050 x 0.396
= 0.198 kgms-1

fourth

im nto sure about this some1 ddouble chceck me

2 x m x v?? <_ ????

fifth average force is f = mv / t

= 0.4 x 3 x3.96 / 0.050

= 31.68 N

ownage.

the one and only

DaZ
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CasedIt
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Report Thread starter 16 years ago
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cheers m8!!!!
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DazYa
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(Original post by CasedIt)
cheers m8!!!!
yahh PM me nxt time
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iiikewldude
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Report 16 years ago
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(Original post by CasedIt)
stuck on momentum q ples help:

A ball of mass 0.050KG rests on some tennis strings and is held horizontally

a i calculate the difference in magnitude between the two forces on the ball when the racket is accelerated upwards at 2ms

( F = ma ? )

b i The ball is dropped from rest at a point 0.80 m above the racket head

ii calculate the speed of the ball before impact

iii calculate the momentum of the ball before impact

iiii change in moment of ball during impact

iv the average force during impact with contact time of 0.050 s
B)ii) V^2 = U^2 + 2as where U=O
V^2 = 2*9.8*0.8 = 15.68
V = Sqrt 15.68 = 4.0 ms^-1

iii)change in momentum = m (delta) v
=0.05*(4.0-0)
=0.2 kgms^-1

iiii) I dunno??

iv) change in momentum = Ft
change in momentum/t = F
0.2/0.05 = 4kgms^-2
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DazYa
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he probably won't get any marks for bringing suvat into it, derive the proper physics formulae
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