Question

A 3000 kg space probe is moving in a gravity free space at a constant velocity of 300 m/s. To change the direction of space probe, rockets have been fired in a direction perpendicular to the direction of initial motion of the space probe, the rocket firing exerts a thrust of 4000 N for 225 s. The space probe will turn by an angle of :

(neglect the mass of the rockets fired)

• A. ${30}^o$
• B. ${60}^o$
• C. ${45}^o$
• D. ${37}^o$

Answer 1

The correct option is C ${45}^o$

Given : $m=3000kg$ $v=300$ $m/s$

Let the angle by which the space probe turns be $\theta$ having momentum $P$

Momentum of the firing rocket $P^{\prime }=Ft=4000\times 225=900000$ N s

Applying conservation of momentum along x direction :

$mv=Pcos\theta$ .............(1)

Applying conservation of momentum along y direction :

$P^{\prime }=Psin\theta$

$900000=Psin\theta$ ..............(2)

Divide (1) form (2), we get $tan\theta =\frac{900000}{mv}$

$tan\theta =\frac{900000}{3000\left(300\right)}=1$ $⟹\theta ={45}^o$