# 1)a body of mass 3 kg is moving along a straight line with a velocity 24m/s.when it is at a point "p" a force of 9 N acts on the body in a direction opposite to its motion.the time after which it will be at "p" again is 2)a body of mass 2kg is moving along +ve X axis with a velocity of 5m/s. now a force of 10√2N is applied at an angle of 45deg with Xaxis.its vel after 3 secs is 3)A ball reaches a rocket at 60m/s along + Xaxis and leaves the rocket in the opposite direction with the same speed.assume the mass of ball is 50gm and the contact time is 0.02secs.the force exerted by the racket on the ball is

1
by aimanakhtar799

2014-10-12T05:06:52+05:30

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1)
u = 24 m/sec   m = 3 kg   Force = 9 N
acceleration = - 9 N/ 3 kg = - 3 m/sec²
S = displacement from point P =  u t + 1/2 a t² = 0

24 t - 1/2 * 3 * t² = 0
t = 0  or   24 - 3t/2 = 0   ie.,  t = 24*2/3 = 16 seconds.

2)
m = 2 kg     u_x = 5 m/s           F = 10√2 N          Ф = 45°
Acceleration along Y axis a_y = F Sin Ф / m = (10 √2) * (1/√2)  / 2 = 5 m/sec²
Acceleration along X axis a_x = F Cos Ф / m =(10 √2) * (1/√2)  / 2 = 5 m/sec²

Velocity along X direction = Vx = u_x  + a t
at t = 3,       Vx  = 5  + 5 * 3 = 20 m/sec

Velocity along Y direction = Vy = u_y + a t
at t = 3,       Vx  = 0 +  5 * 3 = 15 m/sec

Magnitude of the resultant velocity = √ (Vx² + Vy² ) = √(400+225) = 25 m/sec

Direction of the resultant velocity = tan α =  Vy/Vx = 15/20 = 3/4

α = tan⁻¹ 0.75 = 36.87° with x-axis

3)

Change in the momentum of the ball = mass * (final velocity - initial velocity)
= 50/1000  * (- 60 - 60 ) = - 6 kg-m/sec

Force on the ball by the rocket = change in momentum / time duration for the change
= - 6 / 0.02 = - 300 Newtons
Force is in the direction of -ve X direction