NUMERICAL.

A vehicle of mass 30 quintals moving with a speed of 18km/hr collides with another vehicle of mass 90 quintals moving with a speed of 14.4 km/hr. What will be the velocity of each after the collision if they are in opposite directions?

let the initial and final velocities of 30 quintal body be u and v while that of 90 quintal body be u' and v' respectively.

here we shall use the laws of conservation of energy and momentum

so,

mu + m'u' = mv + m'v'..........................................(1)

and

law of conservation of (kinetic) energy

(1/2)mu^{2} + (1/2)m'u'^{2} = (1/2)mv^{2} + (1/2)m'v'^{2}

or

mu^{2} + m'u'^{2} = mv^{2} + m'v'^{2}....................................(2)

now, here

m = 30 quintals = 3000 kg

m' = 90 quintals = 9000 kg

u = 18 km/hr = 5 m/s

u' = 14.4 km/hr = 4 m/s

thus, equation 1 becomes

3000x5 + 9000x4 = 3000v + 9000v'

or dividing by 3000 on both sides

5 + 12 = v + 3v'

or

v + 3v' = 17...........................................................(3)

or v = 17 - 3v'........................................................(4)

now, equation (2) becomes

3000x5^{2} + 9000x4^{2} = 3000v^{2} + 9000v'^{2}

or

25 + 48 = v^{2} + 3v'^{2}

or

v^{2} + 3v'^{2 }= 73..........................................................(5)

now, by substituting the value of v from (4) in (5), we get

(17 - 3v')^{2} + 3v'^{2 }= 73

or

9v'^{2} + 289 - 102v' + 3v'^{2 }= 73

or

12v'^{2} - 102v' + 216 = 0

dividing both sides by 6, we have

2v'^{2} - 17v' + 36 = 0

now, by solving the above quadratic equation given us two values of v'

that is

v' = 4 m/s or v' = 4.5 m/s

we can substitute both value of v; one by one in (3) to get two corresponding value of v

so,

(a)

if v' = 4m/s ; v = 5 m/s

and

(b) if v' = 4.5 m/s = 3.5 m/s

we shall select option (b) to be the correct values as option (a) represents the initial velocities.

So, the final velocities will be

**v = 3.5 m/ and v' = 4.5 m/s**

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