The velocity of the `CM` of a system changes from `vec v_1 = 4 hat im//s to vecv_2 = 3 hatj m//s` during time `Deltat = 2 s`. If the mass of the system is `m = 10 kg`, the constant force acting on the system is :

Velocity of a particle of mass 2 kg change from vecv_(1) =-2hati-2hatjm/s to vecv_(2)=(hati-hatj)m//s after colliding with as plane surface.

Velocity of a particle of mass 2 kg change from vecv_(1) =-2hati-2hatjm/s to vecv_(2)=(hati-hatj)m//s after colliding with as plane surface.

Kg is the unit of mass in the M.K.S. system.

The velocity (in m/s) of an object changes from vecv_(1) = 10 hati + 4hatj + 2hatk "to" vecv_(2) = 4hati + 2hatj + 3hatk in 5 second. Find the magnitude of average acceleration.

The velocity (in m/s) of an object changes from vecv_(1) = 10 hati + 4hatj + 2hatk to vecv_(2) = 4hati + 2hatj + 3hatk in 5 second. Find the magnitude of average acceleration.

The velocity (in m/s) of an object changes from vecv_(1) = 10 hati + 4hatj + 2hatk " to " vecv_(2) = 4hati + 2hatj + 3hatk in 5 second. Find the magnitude of average acceleration.

The velocity of a body of mass 2 kg as a function of t is given by vec v (t) = 2t hat i + t^2 hatj . Find the momentum and the force acting on it, at time t = 2s.

Velocity vectors of three masses 2kg,1kg and 3 kg are vecv_1 = (hati-2hatj+hatk) m/s,vecV_2 = (2hati +2hatj -hatk) m/s and vecv_3 respectively. If velocity vector of center of mass of the system is zero then value of vecv_3 will be