What is the Velocity Ratio in an Elastic Collision?

[lookitzcathy]

Okay, so my homework question is:

5) An object makes an elastic head-on collision with another "target" object which is initially at rest. If the ratio of incident mass over target mass is 0.5333, what is the velocity of the incident object after the collision in multiples of its incident velocity? Give the result with the appropriate sign taking the incident velocity as positive.

Would any of ya'll be kind enough to help me solve this problem?! Thanks a bunches.

 

[Andrew Mason]

Okay, so my homework question is:

5) An object makes an elastic head-on collision with another "target" object which is initially at rest. If the ratio of incident mass over target mass is 0.5333, what is the velocity of the incident object after the collision in multiples of its incident velocity? Give the result with the appropriate sign taking the incident velocity as positive.

Use the fact that in an elastic collision the relative speeds are the same before and after the collision (speed of approach = speed of separation after collision).

$$m_1v_{1i} = m_1v_{1f} + \frac{m_1}{.5333}v_{2f}$$

and:

$$v_{1i} - 0 = v_{1f} - v_{2f}$$

Two equations and two variables should be all you need to express an answer in terms of $v_{1i}$

AM

 

[lookitzcathy]

Use the fact that in an elastic collision the relative speeds are the same before and after the collision (speed of approach = speed of separation after collision).

$$m_1v_{1i} = m_1v_{1f} + \frac{m_1}{.5333}v_{2f}$$

and:

$$v_{1i} - 0 = v_{1f} - v_{2f}$$

Two equations and two variables should be all you need to express an answer in terms of $v_{1i}$

AM

hey... yea, i got that equation too... but I'm still unsure of how i will be able to find the velocity by just knowing the "ratio" of the incident mass over target mass. sigh*

 

[Andrew Mason]

hey... yea, i got that equation too... but I'm still unsure of how i will be able to find the velocity by just knowing the "ratio" of the incident mass over target mass. sigh*

(1)$$m_1v_{1i} = m_1v_{1f} + \frac{m_1}{.5333}v_{2f}$$

and:

(2)$$v_{1i} - 0 = -(v_{1f} - v_{2f})$$ (I forgot the minus sign before)


Substitute expression for v1f from (2) into (1):

$$m_1v_{1i} = m_1(v_{2f} - v_{1i} + \frac{1}{.5333}v_{2f})$$


$$2m_1v_{1i} = m_1v_{2f}(1 + \frac{1}{.5333})$$

$$v_{2f} = v_{1i}\frac{1.0666}{1.5333} = .6956v_{1i}$$

from (2):

$$v_{1f} = (1 - .6956)v_{1i} = .3044v_{1i}$$

AM

 

[lookitzcathy]

o wow i got it... thank you so much!