Question

Two baby buggies are moving toward a head-on collision on level ground. The first buggy has a mass of 12 kg and is moving at a speed of 3 m/s. The second buggy has a mass of 15 kg and is moving at a speed of 5 m/s. The buggies are each equipped with rugged rubber bumpers that insure the collision is perfectly elastic.
Which of the following is closest to the distance between the buggies 3 seconds after the collision? (You may neglect friction.)

10 m
15 m
20 m
25 m
30 m

Answer 1

Answer:

The distance between the buggies is 25 m.

(d) is correct option.

Explanation:

Given that,

Mass of first buggy = 12 kg

Speed of first buggy = 3 m/s

Mass of second buggy = 15 kg

Speed of second buggy = 5 m/s

Time = 3 sec

We need to calculate the final velocity of first buggy

Using formula of velocity

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

Put the value into the formula

$v_{1}=\dfrac{(12-15)\times3}{12+15}+\dfrac{2\times15\times5}{12+15}$

$v_{1}=5.22\ m/s$

We need to calculate the final velocity of second buggy

Using formula of velocity

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

Put the value into the formula

$v_{2}=\dfrac{2\times12\times3}{12+15}-\dfrac{(12-15)\times5}{12+15}$

$v_{2}=3.22\ m/s$

The velocity separation will be 2 m/s.

We need to calculate the distance for first buggy

Using formula of distance

$s=v_{1}\times t$

Put the value into the formula

$s=5.22\times3$

$s=15.66\ m$

We need to calculate the distance for second buggy

Using formula of distance

$s'=v_{2}\times t$

Put the value into the formula

$s'=3.22\times3$

$s'=9.66\ m$

Total distance between both buggies

$S=s+s'$

Put the value into the formula

$S=15.66+9.66$

$S=25\ m$

Hence, The distance between the buggies is 25 m.