Question

A 0.140 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.900 m/s . It has a head-on collision with a 0.297 kg glider that is moving to the left with a speed of 2.25 m/s . Suppose the collision is elastic.

Answer 1

Answer:

(a). The final velocity of the first glider is −2.15 m/s.

(b). The final velocity of the second glider is -1.67 m/s.

Explanation:

Given that,

Mass of glider = 0.140 kg

Speed = 0.900 m/s

Mass of another glider = 0.297 kg

Speed =2 .25 m/s

Suppose, Find the magnitude of the final velocity of the first glider.

Find the magnitude of the final velocity of the second glider.

(a). We need to calculate the final velocity of the first glider

Using formula of collision

$v_{1}=\dfrac{u_{1}(m_{1}-m_{2})+2m_{2}u_{2}}{m_{1}+m_{2}}$

Put the value into the formula

$v_{1}=\dfrac{0.900(0.140+0.297)+2\times(-0.297)\times2.25}{0.140+0.297}$

$v_{1}=-2.15\ m/s$

(b). We need to calculate the final velocity of the second glider

Using formula of collision

$v_{2}=\dfrac{u_{2}(m_{2}-m_{1})+2m_{1}u_{1}}{m_{1}+m_{2}}$

Put the value into the formula

$v_{1}=\dfrac{-2.25(0.297+0.140)+2\times(0.140)\times0.900}{0.140+0.297}$

$v_{1}=-1.67\ m/s$

Hence,  (a). The final velocity of the first glider is −2.15 m/s.

(b). The final velocity of the second glider is -1.67 m/s.

Answer 2

Answer with Explanation:

We are given that

Mass of glider=m=0.14 kg

Initial speed of first glider, u=0.9 m/s

m'=0.297 kg

Initial speed of second glider, u'=-2.25m/s

a.We have to find the magnitude of final velocity of 0.140 kg glider.

The collision is elastic then the final velocity of 0.140 kg glider

$v=\frac{(m-m')u+2m'u'}{m+m'}$

Substitute the values

$v=\frac{(0.140-0.297)\times 0.9+2\times 0.297\times (-2.25)}{0.140+0.297}=-3.38m/s$

Magnitude of final velocity of 0.140 kg glider=3.38 m/s

b.$v'=\frac{u'(m'-m)+2mu}{m+m'}$

$v'=\frac{-2.25(0.297-0.140)+2\times 0.140\times 0.9}{0.140+0.297}=-0.23 m/s$

Hence, the magnitude of final velocity of 0.297 kg glider=0.23 m/s