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# Physics- Velocity and Kinetic Energy?

Two particles of mass m1 = 1.7 kg and m2 = 3.5 kg undergo a one-dimensional head-on collision as shown in the figure below. Their initial velocities along x are v1i = 15 m/s and v2i = - 7.5 m/s. The two particles stick together after the collision (a completely inelastic collision). (Assume to the right as the positive direction.)

a) Find the velocity after the collision. (m/s)

b) How much kinetic energy is lost in the collision? (J)

### 3 Answers

- Anonymous9 years agoFavorite Answer
CONSERVATION of LINEAR MOMENTUM

(m1v1i)i+(m2v2i)=(m1vif)+(m2v2f)

since they the collision is inelastic

v1f=v2f=vf

making the equation

(m1v1i)+(m2v2i)=(m1+m2)vf

(1.7kg)(1.5m/s)+(3.5kg)(-7.5m/s)=(1.7kg + 3.5kg)vf

-23.7kgm/s=5.2kg(vf)

vf=(-23.7kgm/s)/(5.2kg)

***vf=-4.56 m/s***

For the kinetic energy

is the KE (lost) for each particle?

if I got that right,

use the formula:

Just substitute values and use

KE=(1/2)m1v1f-m1v1i----->for KE lost in particle 1

and

KE=(1/2)m2v2f-m2v2i----->for KE lost in particle 2

Hope this helps you!

^^

- badenLv 44 years ago
properly quite lots you have already got the different men explaining to you what to do with this issue. properly in certainty conservation of potential, what does it mean? properly conservation potential not something is lost so for you to use potential potential and convert it into Kinetic potential PE= mgh KE= a million/2mv^2 so first you calculate the potential potential first so utilising the formuela given above: PE= 100X9.81X50 = 49050J and we pronounced in the previous that no potential is lost so we are in a position to transform it into the Kinetic potential 49050= a million/2X 100X v^2 so make v the region and you get the respond 31.3209... rounding it to the smallest sig we get 30J