Direct central impact occurs between a 300N body moving to

[manal950]

Hi all

1 )
Direct central impact occurs between a 300N body moving to the right with velocity of 6 m/s and 150 N body moving to the left with velocity of 10 m/s . Find the velocity of each body after impact if the coefficient of restitution is 0.8 .

my answer :

Given [/color]::
m1 = 300 /9.81 = 30.58 kg
u1 = 6 m/s
m2 = 150/9.81 = 15.2 kg
u2 = -10 m/s
e = 0.8
V 1 ? v2 ?


m1u1+m2u2 = m1v1+m2v2
30.58X 6 + 15.2 X (-10) = 30.58v1+ 15.2 v2
183.48 -152 = 30.58v1+ 15.2 v2
31.48 = 30.58v1+ 15.2 v2 ===== this is first equation (1)[/color]

e(u1 - u2) = (v2 - v1)
0.8(6 + 10) = v2 - v1
4.8 + 8 = v2 - v1
12.8 = v2 - v1
v2 = 12.8 + v1 ===== this is first equation (2) [/color]

make v2 in fist equation
31.48 = 30.58v1+ 15.2 (12.8 + v1)
31.48 = 30.58v1+ 194.56 + 15.2 v1)
31.48 - 194.56= 30.58v1 + 15.2 v1
-163.08 = 45.78
v1 = -3.6 m/s [/color]

to find v2 = 12.8 + (-3.6) = 9.2 m/s[/color]

 

[Infinitum]

Hi manal950!

Were you unsure you solved it correctly? Your approach looks correct to me.

 

[manal950]

thanks so much ..