A 0.200 kg plastic ball moves with a velocity of 0.30 m/s. It collides with a second plastic ball of mass...?

Question

A 0.200 kg plastic ball moves with a velocity of 0.30 m/s. It collides with a second plastic ball of mass of 0.100 kg, which is moving along the same line at a speed of 0.10 m/s. After the collision, both balls continue moving in the same, original direction. The speed of the 0.100 kg ball is 0.26 m/s. What is the new velocity of the 0.200 kg ball?

Please help me with this problem.
I'm having trouble understanding it.

Answer 1

ok so. think of it this way.

there is a football player, he is lets say bigger.[the .2kg ball] then there is another one, he is smaller.[the .1kg ball] you know that

find- Vf of the .2kg ball

know- m[1] is .2kg
m[2] is .1kg
vi[1] is .3m/s
vi[2] is .1m/s
vf[2] .26m/s
Conservation of Momentum: m1v1 = m2 v2

so v1 = .13 m/s..

goodluck. i hate physics and suck at it!

Answer 2

m1 = 0.2kg
m2 = 0.1kg
v10 = 0.3 m/s
v20 = 0.1 m/s
v2f = 0.26 m/s
v1f = ?

Use Conservation of momemtum

m1v10 + m2v20 = m1v1f + m2v2f,

and solve for v1f:

v1f = (m1v10 + m2v20 - m2v2f)/m1 = 0.22 m/s

This is different than the above answers because I did not assume a totally inelastic collision. There was no mention of the two bodies sticking together, just that they were moving in the same direction before and after the collision.

By the way, the collision does not conserve mechanical energy. You will find that about 128 mJoules of kinetic energy is lost during the collision, to other forms of energy such as heat, sound, etc.

Answer 3

k here is a diagram if you dont understand:

BALL 1 -----------> Ball2--->

Conservation of Momentum:
m1v1 = m2 v2

You are given that the velocityof small ball is .26 m/s* .1 kg = .026
so m1v1 = .026, which means .2 v1=.026
.026/.2

v1 = .13 m/s..
This problem is tricky, because since this is an inellastic collision, the initial speeds are not needed, i.e. some energy is converted to sound and heat, so there is no way to calculate speed of ball by using law of energy.. so you use law of momentum.. Hope this helps

Answer 4

The momentum (mass x velocity) within the system of two moving balls will be unchanged. The .1kg ball GAINED .16 m/sec. Right away you know the .200 (twice the mass) ball will have lost 1/2 the speed the lighter ball gained or .08 m/sec leaving it with a velocity of .22m/sec.
Momentum before collision = (.2kg x .3m/s) + (.1kg x .1m/s)
= .06k-m/s + .01k-m/s = .07k-m/s
Momentum after colision = (.2kg x .22m/s) + (.1kg x .26m/s)
= .044k-m/sec + .026k-m/s = .07k-m/s

Answer 5

Conservation of momentum allways applies. Conservation of energy only applies to perfectly elastic collisions. It is likely that two plastic balls will undergo an inelastic collision. So solve using conservation of momentum. (As a check you can confirm that the Kinetic energy in the system after the collision goes down. )

Answer 6

Matt & S. both started with the wrong equation.
vejj got it exactly right; I gave him my vote.
My way of explaining it:

∆momentum is the same for each and equal to ∆V2*m2 = (.26-.1)*.100 = .016. Dividing this by m1 (=.200) gives ∆V1 = .08
So, V1f = .30-.08 = .22 m/s

Answer 7

Conservation of momentum
Initial momentum sum = Final momentum sum
m1 x u1 + m2 x u2 = m1' x u1' + m2' x u2' ....... (1)

m1 = m1' = 0.200 kg
m2 = m2' = 0.100 kg
u1 = 0.30 m/s
u2 = 0.10 m/s
u2' = 0.26 m/s
u1' = ?

Plug the values in equation (1)
Then solve for u1'