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.
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Answer:
Given -
[tex] \\ \sf \: 90 \times \frac{5}{18} = 25 \: m/s \\ [/tex]
[tex] \\ \sf \: 18 \times \frac{5}{18} = 5 \: m/s \\ [/tex]
Change in momentum -
[tex] \\ \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. \\ [/tex]
Acceleration -
[tex] \\ \sf \: v = u + at \: \\ \\ \\ \implies \sf \: 5 = 25 + a \times 4 \\ \\ \\ \implies \sf \: - 5 \: m/ {s}^{2} \\ [/tex]
Magnitude of the force required -
[tex] \\ \sf \: |F| = m |a| \\ \\ \\ \implies \sf \: |F| = 1200 \times 5 \\ \\ \\ \implies \sf \: |F| = 6000 \: N\\ [/tex]
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.