Question

A motor car of mass 1200 kg is moving along a straight line with uniform velocity of 90 km/h. The velocity slows down to 12 km/h in 4 sec by an unbalanced force. Calculate the acceleration and change in
momentum. Also calculate the magnitude of force.​

Answer 1

Answer:

6000N

Explanation:

Initial velocity of the car, u=90 km/h=25 ms−1; 

Final velocity of the car, v=18 km/h=5 ms−1; 

Time, t=4 s;

v=u+at

5=25+(a×4)

∴−a×4=20 or a=4−20=−5ms−2

Change in momentum, Δp=m(v−u)

                                              =1200(5−25)=1200×(−20)=−24000Ns

Magnitude of force

Therefore,  the force required, F=tm(v−u)=4−24000=−6000N

Answer 2

Answer:

Explanation:

Given,

Mass of the motor car, m = 1200 kg

Initial velocity of motor car, u = 90 km/h = 90 x 5/18 = 25 m/s

Final velocity of motor car, v = 18 km/h = 18 x 5/18 = 5 m/s

Time taken by motor car, t = 4 seconds

To Find,

Acceleration = ?

Change in momentum = ?

Force = ?

Formula to be used,

First equation of motion, v = u + at

Change in momentum = m (v - u)

Force = ma

Soloution,

Putting all the values, we get

v = u + at

⇒ 5 = 25 + a × 4

⇒ 5 - 25 = 4a

⇒ - 20 = 4a

⇒ - 20/4 = a

⇒ a = - 5 m/s².

Hence, The Acceleration of the motor car is - 5 m/s².

Now, the Change in momentum,

Change in momentum = m (v - u)

⇒ Change in momentum = 1200 × (5 - 25)

⇒ Change in momentum = 1200 × (- 20)

⇒ Change in momentum = - 24000 kg m/s

Hence, the Change in momentum is  - 24000 kg m/s.

Now, the force required to decrease the velocity,

Force = ma

⇒ Force = 1200 × (- 5)

⇒ Force = - 6000 N

Hence, the force required to decrease the velocity is - 6000 N.