Quetions of body under collison

[manal950]

Hi all

http://store2.up-00.com/June12/pG127901.jpg


V^2 = u^2 + 2as
V = 8.85 m/s which will be intial veloctiy befor hits the second body

Bu using
e(u1 - u2 )=V2 -V1
m1u1 +m2u2 =m1v1+m2v2



for the body which is 20 N
mass 20.3
u1 = 8.86 0m/s
V1 after collision = 0 m/s
for the second body which is 40 N
mass = 4.077
u2 = 0
v2 we will find final velocity after collion
m1u1 +m2u2 =m1v1+m2v2
2.03 X 8.86 + 4.077 X 0 = 2.03X(0) + 4.077V2

 

[Xisune]

v1 after the collision is not 0, this is because the objects have different masses.
But since u2 is 0, you can solve for v1 using:

V = (m1-m2)u/(m1 + m2)

After solving for v, using that for kinetic energy and set that equal to potential energy.

 

[manal950]

you mean by ""using that for kinetic energy and set that equal to potential energy""

this equation
m1u1 +m2u2 =m1v1+m2v2

and is the final answer for height
1.78 and 1.44 m

 

[manal950]

Because we have 2 unknown V1 and V2 I will find it by this two equation ...

m1u1 +m2u2 = m1v1 + m2V2
2.03X8.8+4.077(0) = 2.03(v1)+4.077V2
17.6204 = 2.03V1 + 4.077V2 ====== fist equation (1)

e(u1-u2) = (V2 - V1)
0.8(8.86 - 0 ) V2 - V 1
V2 = 7.088 + V 1 this second equation (2)
so I will make V2 in first equation
17.6204 = 2.03V1 + 4.077(7.088 + V1 )
17.6204 = 2.03V1 + 28.89+4.077V1
V1 = -1.84m/s
Now I will findV2
17.06204 = 2.03(-1.84) + 4.077V2
21.6551 = 4.077V2
V2 = 5.24 m/s

so is that correct or not ??

 

[Xisune]

you mean by ""using that for kinetic energy and set that equal to potential energy""

this equation
m1u1 +m2u2 =m1v1+m2v2

and is the final answer for height
1.78 and 1.44 m

I actually gave you the wrong equation, my mistake.

The correct equation is:

v2 = (2m1 u1)/(m1+m2)

Solve for v2, then use that value for kinetic energy and set that equal to potential energy, which is mgh.

 

[manal950]

see my post number 4

 

[Xisune]

Because we have 2 unknown V1 and V2 I will find it by this two equation ...

m1u1 +m2u2 = m1v1 + m2V2
2.03X8.8+4.077(0) = 2.03(v1)+4.077V2
17.6204 = 2.03V1 + 4.077V2 ====== fist equation (1)

e(u1-u2) = (V2 - V1)
0.8(8.86 - 0 ) V2 - V 1
V2 = 7.088 + V 1 this second equation (2)
so I will make V2 in first equation
17.6204 = 2.03V1 + 4.077(7.088 + V1 )
17.6204 = 2.03V1 + 28.89+4.077V1
V1 = -1.84m/s
Now I will findV2
17.06204 = 2.03(-1.84) + 4.077V2
21.6551 = 4.077V2
V2 = 5.24 m/s

so is that correct or not ??

Yes, the method is correct, but the velocity is off by a bit.

 

[manal950]

m1u1 +m2u2 = m1v1 + m2V2
2.038735X8.85889+4.07747(0) = 2.038735(v1)+4.07747V2
18.0609 = 2.038735(v1) + 4.07747V2 ====== first equation (1)

e(u1-u2) = (V2 - V1)
0.8(8.85889 - 0 ) V2 - V 1
V2 = 7.087112 + V 1 this second equation (2)
so I will make V2 in first equation
18.0609 = 2.038735(v1) + 4.07747(7.087112 + V 1)
18.0609 = 2.038735(v1) + 28.89748 + 4.07747V1
18.0609 - 28.89748 = 2.038735(v1) + 4.07747V1
-10.83658 = 6.116205 V1
V1 = -1.77178 m/s

Now I will findV2
V2 = 7.087112 + (-1.77178)
V2 = 5.3153 m/s

----------- now for the e = 1 I will get velocity V2 -----------------
e(u1-u2) = (V2 - V1)
1(8.85889 - 0 ) =V2 - V1
V2 = 8.85889 + V 1
I will make V2 in 1
18.0609 = 2.038735(v1) + 4.07747(8.85889 + V 1)
18.0609 - 36.12185 = 6.116205V1
V1 = -2.9529 m/s
V2 = 8.85889 + (-2.9529)
v2 = 5.9059 m/s Now finding the height :
when e = 1
1/2mv2 = mhg
0.5(2.03873)(5.9059)^2 = 2.03873 X 9.81 X h
h = 1.7 m

when e = 0.8
1/2mv2 = mgh
0.5(2.03873)(5.3153)^2 = (2.03873)gh
m = 1.4 m