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