Question

Q8. A vehicle of mass 3000 kg is moving along a straight line with a uniform velocity of 80km/h. Its velocity is
slowed down to 50km/h in 5s by an unbalanced external force. Calculate the acceleration and change in
momentum and net force.​

Answer 1

Mass of vehicle, m = 3000 kg

Initial velocity of vehicle, u = 80km/h

Final velocity of vehicle, v = 50km/h

Time taken to bring about the change, t = 5s

• TO FIND ACCELERATION:

Acceleration is the change in velocity per unit time, which is given by:

⇒ a = (v - u) / t

Where, u & v represents initial and final velocity respectively.

Now, the velocities are given in km/h and the time taken to bring about this change is given in s. So, Let's convert the velocities into m/s by multiplying them by 5/18.

• u = 80 × 5/18 = 200/9 m/s
• v = 50 × 5/18 = 125/9 m/s

Now,

⇒ a = (125/9 - 200/9) / 5

⇒ a = (125 - 200) / 45

⇒ a = -5/3 m/s² or -1.67 m/s²

Note: The negative sign here signifies that the magnitude of velocity is getting reduced.

• TO FIND CHANGE IN MOMENTUM

Momentum is given by,

• p = mv

Here, The initial momentum is mu and the final momentum is mv, so, change in momentum is given by,

⇒ ∆p = mv - mu

⇒ ∆p = m(v - u)

⇒ ∆p = 3000 (50 - 80)

⇒ ∆p = 3000 × -30

⇒ ∆p = -9000 kg.m/s

• TO FIND NET FORCE

We know, Force is the change in momentum per unit time and is given by,

⇒ F = ∆p / ∆t

We found the change in momentum in the previous section to be -9000 kg.m/s , so

⇒ F = -9000/5

⇒ F = -1800 N

Note: The negative sign here denoted that the force is acting against the motion of the vehicle.

Answer 2

Given :-

Mass 3000 kg is moving along a straight line with a uniform velocity of 80km/h. Its velocity is  slowed down to 50km/h in 5s by an unbalanced external force

To Find :-

Acceleration

Momentum

Net force

Solution :-

We know that

v = u + at

Since velocity is in km/h

For initial velocity

80 = 80  × 5/18 = 400/18 = 200/9

For final velocity

50 = 50 × 5/18 = 250/18 = 125/9

$\sf\dfrac{125}{9} = \dfrac{200}{9} + a(5)\;$

$\sf \dfrac{125 - 200}{9} = 5a$

$\sf \dfrac{-75}{9} = 5a$

$\dfrac{-25}{3} = \sf 5a$

$\sf -25 = 3(5a)$

$\sf -25 = 15a$

$\sf\dfrac{-25}{15}=a$

$\sf\dfrac{-5}{3}=a$

For Momentum

$\sf\triangle p = m(v-u)$

$\sf\triangle p = 3000(50-80)$

$\sf\triangle p = 3000(-30)$

$\sf\triangle p = -9000\; kg m/s$

For net force

$\sf F_n = \dfrac{Momentum}{Time}$

$\sf F_n = \dfrac{-9000}{5}$

$\sf F_n = -1800 \; N$