Question

A motorcar of mass 1200 kg is moving along a straight line
with a uniform velocity of 90 km/h. Its velocity is slowed down
to 18 km/h in 4 s by an unbalanced external force. Calculate
the acceleration and change in momentum. Also calculate the
magnitude of the force required.​

Answer 1

Answer:

Given -

• Mass of the motor car = 1200 kg.
• Initial velocity, u = 90 km/ h .

$\\ \sf \: 90 \times \frac{5}{18} = 25 \: m/s$

• Final velocity, v = 18 km/h.

$\\ \sf \: 18 \times \frac{5}{18} = 5 \: m/s$

• Time taken, t = 4s.

Change in momentum -

$\\ \sf \Delta \: P = mv - mu \\ \\ \\ \footnotesize \dag \: \: \sf \underline{now \: \: by \: \: putting \: \: the \: \: given \: \: values \: - } \\ \\ \\ \implies \sf \Delta \: P = m \times 5 - m \times 25 \\ \\ \\ \implies \sf \: \Delta \: P = - 24000 \: kg \: \: m \: per \: sec.$

Acceleration -

$\\ \sf \: v = u + at \: \\ \\ \\ \implies \sf \: 5 = 25 + a \times 4 \\ \\ \\ \implies \sf \: - 5 \: m/ {s}^{2}$

Magnitude of the force required -

$\\ \sf \: |F| = m |a| \\ \\ \\ \implies \sf \: |F| = 1200 \times 5 \\ \\ \\ \implies \sf \: |F| = 6000 \: N$

Answer 2

Explanation:

From above data we have m is 1200 kg, u is 90 km/hr or 25 m/s, v is 18 km/hr or 5 m/s and t is 4 sec.

Using the first equation of motion,

v = u + at

→ 5 = 25 + a(4)

→ -20 = 4a

→ a = -5

(Negative sign retardation)

Hence, the acceleration of the motorcar is 5 m/s².

Now,

Change in momentum = mv - mu

= 1200(5 - 25)

= 1200(-20)

= -24000 kg m/s

Force is defined as the product of mass and acceleration.

F = ma

Substitute the values,

→ F = 1200 |-5|

→ F = 1200 × 5

→ F = 6000 N

Hence, the force required is 6000N.