physics

[ruby92]

ahh...

given momentum and time how do u find force?

[A.T]

F=ma
mv= momentum

change in momentum per unit time is force.
change in momentum is (mv-mu) = m(v-u)

Hence F=(m(v-u))/t
as (v-u)/t = a , we get F=ma

Dunno if this anwers your question LOL

[ruby92]

mass is 12kg at u=5ms-1 collides with another obj at rest with mass of 8

i found out that the V total velocity is 3
and that kinetic energy lost is -60J
how do we find the exchange force..which By the way is 73N

wht exactly are the differences in momentum and kinetic energy when objects collide elastically and inelastically
 

[Monica]

@A.T : Thank you for your help in the forum, I can see you are new so welcome. +rep.

[astarmathsandphysics]

elastic collisions - momentum and energy conserved
inelastic collisions momentum conserved, energy not conserved

[ruby92]

a nucleus of mass 1.67*10^-27 kg moving with a velocity of 2*10^4 collides head on with a boron nucleus of mass 1.7*10^-27kg orginally at rest.
find the velocity of the boron nucleus if the collison is perfectly elastic?

[A.T]

a nucleus of mass 1.67*10^-27 kg moving with a velocity of 2*10^4 collides head on with a boron nucleus of mass 1.7*10^-27kg orginally at rest.
find the velocity of the boron nucleus if the collison is perfectly elastic?

Data you know:
m1=1.67*10^-27 kg
m2=1.7*10^-27 kg
u1=2*10^4 ms-1
u2=0
In elastic the total momentum(mv) and the energy(1/2 mv^2) is conserved.
For conserved momentum u have eq:(denote as eq 1 )
m1u1+ m2u2 =  m1v1 + m2v2
And for energy conserved the eq is: (denote as eq 2)
(1/2) m1 u1² + (1/2) m2 u2² = (1/2) m1 v1² + (1/2) m2 v2²

now substitute given values in both eq (1 and 2) and solve them simultaneously and u'll get values of v1 and v2. v2 will be the velocity of the boron nucleus.
this link can help u for your question: http://answers.yahoo.com/question/index?qid=20081104122100AABK0SO

If collisions were INelastic then:(Just for ur concept)
m1=1.67*10^-27 kg
m2=1.7*10^-27 kg
u1=2*10^4 ms-1
u2=0
Using m1u1+ m2u2 =  m1v1 + m2v2
as the collision is perfectly elastic the final velocity of both objects after collision would be zero,and knowing that u2=0 we get:
m1u1=V(m1+m2)
put the given values and you will get v = 9910 ms-1 (correct to 3 significant fig.)

[ruby92]

ok a mass of 8kg moving at 6ms-1 towards another block of mass 4kg at 3ms-1
its a head on collison
how would u find the momentum transferred from one block to the other. if its elastic? and if its inelastic?

tnkx 

[astarmathsandphysics]

For inelastic you would have to know the velocity of one arter the collision then do m(v-u) for inelastic use restitution principle.

[A.T]

ok a mass of 8kg moving at 6ms-1 towards another block of mass 4kg at 3ms-1
its a head on collison
how would u find the momentum transferred from one block to the other. if its elastic? and if its inelastic?

tnkx  

If inelastic: (they would become one particle(join) after collision and will have the same velocity)
formula for inelastic:
m1u1 + m2u2 = V(m1+m2)
Use it to find V(the combined velocity) multiply it by masses to get the momentum.

For elastic u'll have to do all the long working shown in last last question,but as far as i know in A-levels they give u a value of one of the final velocity which makes it easier and u can simply find it using
m1u1+ m2u2 =  m1v1 + m2v2

[astarmathsandphysics]

http://www.astarmathsandphysics.com/a_level_maths_notes/M1/a_level_maths_notes_m1_impulse.html

You do need the speed of one of the particles after the collision, or a condition - the become one particle so have the same speed, one has twice the speed of the other, one particle comes to rest etc so you can use the formula Impulse=m(v-u)

[ruby92]

A 30 kg block (M1) is placed on a frictionless plane that inclines at a 30° angle with respect to the surface of Earth. This block is connected to another 20 kg block (M2) via a weightless rope over a frictionless, ideal pulley. The second block is hanging vertically.
What is the tension of the rope?

(a) 55.0 N
(b) 92.4 N
(c) 147.4 N
(d) 176.6 N
(e) 294.6 N

explained as simple as possible 

[ruby92]

A 10-kg sign is suspended by three ropes, each supporting an equal portion of the sign's weight. The two end ropes make an angle of 70º to the horizontal. What is the tension on each of the ropes?

could someone please solve this fast

[A.T]

A 30 kg block (M1) is placed on a frictionless plane that inclines at a 30° angle with respect to the surface of Earth. This block is connected to another 20 kg block (M2) via a weightless rope over a frictionless, ideal pulley. The second block is hanging vertically.
What is the tension of the rope?

(a) 55.0 N
(b) 92.4 N
(c) 147.4 N
(d) 176.6 N
(e) 294.6 N

explained as simple as possible  

Resolving forces on the 30 kg mass:
T-30gsin(30)=ma
T-30gsin(30)=30a ....1

Resolving forces on 20 kg mass:
W-T=ma
20g-T=20a.....2

solving 1 and 2 simulataneously:
T-30gsin(30) + 20g-T = 30a + 20a
a=0.981 ms-2

T=20g-20a from 2
T=20*9.81 - 21(.981)
T=175.599=175.6 N

Hence answer is C

A 10-kg sign is suspended by three ropes, each supporting an equal portion of the sign's weight. The two end ropes make an angle of 70º to the horizontal. What is the tension on each of the ropes?

could someone please solve this fast

According to question it seems the center rope in vertical.
IF all ropes were vertical the tension in each string would be:
3T=W
3T=10g
T=100/3 N

Hence the center rope has a tension of 100/3 N

now for the end ropes, as they are at 70 degree angle both their vertical component is:
Tsin(70)=100/3
T=35.5N
hence tension in the both end rope is 35.5N

[ruby92]

c are u sure?
isnt it d??