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Q15:
An object of mass 100 kg is accelerated uniformly from a velocity of 5 $m s^{-1}$ to 8 $m s^{-1}$ in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Solution :

Step 1: Identifying Given Parameters

Based on the problem statement, we define the following physical quantities:

  • Mass ($m$) = $100\text{ kg}$
  • Initial velocity ($u$) = $5\text{ m s}^{-1}$
  • Final velocity ($v$) = $8\text{ m s}^{-1}$
  • Time interval ($t$) = $6\text{ s}$

Step 2: Calculation of Initial and Final Momentum

Momentum ($p$) is defined as the product of mass and velocity [$p = m \times v$].

Initial Momentum ($p_i$):

$p_i = m \times u$

$p_i = 100\text{ kg} \times 5\text{ m s}^{-1} = 500\text{ kg m s}^{-1}$

Final Momentum ($p_f$):

$p_f = m \times v$

$p_f = 100\text{ kg} \times 8\text{ m s}^{-1} = 800\text{ kg m s}^{-1}$

Step 3: Calculating Force Exerted

According to Newton's Second Law of Motion, force is the rate of change of momentum [$F = \frac{\Delta p}{t}$] or alternatively [$F = m \times a$].

Calculating Acceleration ($a$):

Using the first equation of motion [$a = \frac{v - u}{t}$]:

$a = \frac{8\text{ m s}^{-1} - 5\text{ m s}^{-1}}{6\text{ s}}$

$a = \frac{3\text{ m s}^{-1}}{6\text{ s}} = 0.5\text{ m s}^{-2}$

Calculating Magnitude of Force ($F$):

$F = m \times a$

$F = 100\text{ kg} \times 0.5\text{ m s}^{-2}$

$F = 50\text{ N}$

Force Application (Direction of Acceleration) m=100kg u=5 m/s v=8 m/s

Step 4: Summary Table

Quantity Result
Initial Momentum ($p_i$) $500\text{ kg m s}^{-1}$
Final Momentum ($p_f$) $800\text{ kg m s}^{-1}$
Force ($F$) $50\text{ N}$

Final Solution: The initial momentum is $500\text{ kg m s}^{-1}$, the final momentum is $800\text{ kg m s}^{-1}$, and the magnitude of the force exerted on the object is $50\text{ N}$.


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