**Find out their common velocity:**

An aeroplane weighing 40 kN horizontally moves with a velocity of 20 m/sec (72 kmph, that means 72 km per hour). A helicopter of weight 20 kN is proceeding at 30 m/sec (108 kmph) towards North in the same horizontal plane & collides with aeroplane. If the two masses get entangled after collision, find out their common velocity.

**Solution**

As momentum is a vector along the direction of the velocity of the mass, while the momenta M_{1} and M_{2} of the aeroplane and the helicopter, respectively, are along two different directions of the two masses, vector rules have to be adopted to determine its resultant momentum.

M¯_{1} =(Momentum of aeroplane before collision in units (kg m/sec) along )

= (40 × 20 )/ g × 1000

M¯ _{2} = (Momentum of helicopter before collision alongin (kg m/sec))

= (20 × 30)/ g × 1000

The combined mass = (60 × 1000)/ g kg

Therefore, if M_{ c} is the resultant momentum along direction at ∠ α with OA then

V¯c is the common velocity along

∴ M¯ c = M¯_{1} + M¯ _{2}

1000 × (60/g) V¯_{c} = [(40 × 20 )/g +( 20 × 30)/g (OB ↑)]× 1000

This is a vector addition,

∴ Vc = 100 /6= 16.67 m / sec along OR

where,

tan α = AR/ OA = 600 /800 = 0.75

∴ α = 36^{o}.87