Speed of first swallow is 20 m/s, speed of second swallow is 15 m/s, mass of first swallow is 0.270 kg, mass of coconut of first swallow is 0.80 kg, mass of second swallow is 0.220 kg, mass of coconut of second swallow is 0.70 kg, speed of first coconut after collision is 13 m/s at 10° south-west direction, speed of second coconut after collision is 14 m/s at 30° north-east direction.
The free body diagram as per the data is shown below in figure 1.
Write the equation for momentum conservation in x-direction for the collision of birds.
m1v1x+m2v2x=m3v3x+m4v4x+m5v5x
Here,m1 is the mass of first swallow, m2 is the mass of second swallow, m3 is the mass of coconut of first swallow, m4 is the mass of coconut of second swallow, m5 is the mass of tangled-up swallows, v1x is the x-component of velocity of first swallow, v2x is the x-component of velocity of second swallow, v3x is the x-component of velocity of coconut of first swallow, v4x is the x-component of velocity of coconut of second swallow, and v5x is the x-component of velocity of birds just after the collision.
Rewrite the above equation in terms of v5x by substituting 0 m/s for v1x and v2x.
m1(0 m/s)+m2(0 m/s)=m3v3x+m4v4x+m5v5xv5x=−m3v3x+m4v4xm5 (I)
Write the equation for momentum conservation in y-direction for the collision of birds.
m1v1+m2v2=m3v3y+m4v4y+m5v5y
Here, v1 is the velocity of first swallow, v2 is the velocity of second swallow, v3y is the y-component of velocity of coconut of first swallow, v4y is the y-component of velocity of coconut of second swallow, and v5y is the y-component of velocity of birds just after the collision.
m1v1+m2v2=m3v3y+m4v4y+m5v5yv5y=m1v1+m2v2−m3v3y−m4v4ym5 (II)
Write the equation to find the magnitude of velocity of birds just after the collision.
v=v5x2+v5y2
Here,v is the magnitude of velocity of birds just after the collision.
Rewrite the above equation by substituting equation (I) and (II).
v=(−m3v3x+m4v4xm5)2+(m1v1+m2v2−m3v3y−m4v4ym5)2 (III)
Write the equation to find the direction of v.
θ=tan−1(v5yv5x)
Here,θ is the direction of v.
Rewrite the above equation by substituting equations (I) and (II).
θ=tan−1((m1v1+m2v2−m3v3y−m4v4ym5)(−m3v3x+m4v4xm5))=tan−1(m1v1+m2v2−m3v3y−m4v4ym3v3x+m4v4x) (IV)
Conclusion:
Substitute 0.80 kg for m3, (13 m/s)(cos260°) for v3x, 0.70 kg for m4, (14 m/s)(cos60°) for v4x, 1.07 kg for m1, 0.92 kg for m2, 20 m/s for v1,−15 m/s for v2, (13 m/s)(sin260°) for v3y,(1.07 kg+0.92 kg) for m5, and (14 m/s)(sin60°) for m4y in equation (III) to find v5.
v=(−(0.80 kg)((13 m/s)(cos260°))+(0.70 kg)((14 m/s)(cos60°))(1.07 kg+0.92 kg))2+((1.07 kg)(20 m/s)+(0.92 kg)(−15 m/s)(1.07 kg+0.92 kg))2−((0.80 kg)((13 m/s)(sin260°))+(0.70 kg)((14 m/s)(sin60°))(1.07 kg+0.92 kg))2=400 m/s=20 m/s
Substitute 0.80 kg for m3, (13 m/s)(cos260°) for v3x, 0.70 kg for m4, (14 m/s)(cos60°) for v4x, 1.07 kg for m1, 0.92 kg for m2, 20 m/s for v1,−15 m/s for v2, (13 m/s)(sin260°) for v3y,(1.07 kg+0.92 kg) for m5, and (14 m/s)(sin60°) for m4y in equation (IV) to find θ.
θ=tan−1((1.07 kg)(20 m/s)+(0.92 kg)(−15 m/s)−(0.80 kg)((13 m/s)(sin260°))−(0.70 kg)((14 m/s)(sin60°))(0.80 kg)((13 m/s)(cos260°))+(0.70 kg)((14 m/s)(cos60°)))=tan−1(−3.077)=−72°
Negative value of tan function indicates that the angle lies in second quadrant. That is, the above angle is measured with respect to negative x axis in clockwise direction.
Write the equation to find the angle made by v5 with respect to positive x-axis in counterclockwise direction.
ϕ=90°−|θ|
Here, ϕ is the angle made by v5 with respect to positive y-axis in counterclockwise direction.
Substitute −72° for θ in the above equation to find ϕ.
ϕ=90°+|−72°|=18°
Therefore, the velocity is 20 m/s at 20° north-west direction.