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.
As momentum is a vector along the direction of the velocity of the mass, while the momenta M1 and M2 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
tan α = AR/ OA = 600 /800 = 0.75
∴ α = 36o.87