Question

Particle a moves along the line

$y = 4 \sqrt{3} m$
with constant velocity vector
of magnitude 2 m/s and directed parallel to the positive x-axis.particle b starts at the
origin with zero speed and constant acceleration a (of magnitude 4m/s^2)at the same instant that the particle
a passes the y-axis. the angle
$\alpha$
between a and the positive y-axis that would result in a collision between these two particles should have a val-
ue equal to​

Answer 1

Given:

Motion of particle A is along the line y = 4√3 m

Velocity of A = 2 m/s

Initial speed of particle B = 0 m/s

Constant acceleration of B = 4 m/s²

Angle between particle A and the y-axis =α

To find:

The angle α.

Solution:

If the two particles collide then their x and y motion must be equated separately and must coincide.

Motion of particle B in y axis:

S = 1/2*a(y)*t² = 1/2*4*cosα*t² = 2*cosα*t²...........(1)

The motion of particle is along y = 4√3 m............(2)

Equating both these equations:

4√3 = 2t² cosα

For motion in x axis, both the particles must coincide, therefore:

For particle B:

S = 1/2*a(x)*t² = 1/2*4*sinα*t² = 2*sinα*t².............(3)

For particle A:

S = vt = 2t...............(4)

Equating equation 3 and 4 we get:

2t = 2t² sinα

t = 1/ sinα

Putting the value of t in 4√3 = 2t² cosα we get:

2√3 = cosα/ sin²α

2√3 = cosα/ 1-cos²α

1-cos²α = cosα/2√3

2√3 - 2√3cos²α- cosα = 0

2√3cos²α + cosα - 2√3 = 0

Using the quadratic formula to find the roots of the equation we get:

-1±√(1+4*2√3*2√3) / 4√3

-1±√(1 + 48) / 4√3

-1 ±√49 / 4√3

-1±7/ 4√3

6/4√3 = √3/2 and 8/4√3 = 2/√3 (not valid as it is greater than 1 and cosα lies between -1 and 1)

cosα comes out to be √3/2, therefore α = 30°

The value of α is 30°.