Question

The velocity of the CM of a system
changes from v1= 4îm/s to v2 = 3jm/s
during time ∆t= 2s. If the mass of the
system is m=10 kg, the constant force
acting on the system is:​

Answer 1

Given:

The velocity of center of mass changes from $v_{1} = 4i \frac{m}{s}$ to $v_{2} = 3j \frac{m}{s}$ in 2 seconds.

Mass of the system = 10 kg

To Find:

Constant force acting on the system

Solution:

Second Law of motion:

The second law states that the rate of change of momentum of a body is directly proportional to the force applied.

$F = \frac{dM}{dt}$

Initial momentum = $M_{1} = mv_{1}$

Final momentum = $M_{2} = m v_{2}$

$F = \frac{m (v_{1} - v_{2} )}{t_{2} -t_{1} }$

put the velocities in their vector form.

$F = \frac{10(4i - 3j)}{2} \\F = 20i - 15j$

Constant force acing on the system is $20i - 15j$ .