Laws of Motion Quiz - ICSE Class 9 Physics
Free practice quiz: Laws of Motion (ICSE Class 9 Physics). 51 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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A body of mass $m$ is moving with velocity $v$. If its velocity is doubled, what will be its new kinetic energy?
- Half of the original kinetic energy
- The same as the original kinetic energy
- Twice the original kinetic energy
- Four times the original kinetic energy
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Answer: D. Four times the original kinetic energy
The kinetic energy of a body is given by the formula $KE = \frac{1}{2}mv^2$. If the velocity is doubled, the new velocity is $2v$. The new kinetic energy will be $KE_{new} = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2) = 4 \times \frac{1}{2}mv^2 = 4 \times KE$. Therefore, the new kinetic energy is four times the original kinetic energy.
A body of mass 10 kg is moving with a velocity of 5 m/s. If an additional force of 20 N is applied in the direction of motion for 4 seconds, what will be the final velocity?
- 7 m/s
- 9 m/s
- 11 m/s
- 13 m/s
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Answer: D. 13 m/s
First, let's find the change in momentum due to the applied force. Impulse = Force $\times$ time $= 20 \text{ N} \times 4 \text{ s} = 80$ N-s. The change in momentum is also equal to $m \times \Delta v$. So, $80 = 10 \text{ kg} \times \Delta v$. This gives $\Delta v = \frac{80}{10} = 8$ m/s. The initial velocity is 5 m/s. Since the force is applied in the direction of motion, the velocity will increase. The final velocity $v_{final} = v_{initial} + \Delta v = 5 \text{ m/s} + 8 \text{ m/s} = 13$ m/s.
A force of 100 N acts on a body of mass 50 kg. If the body starts from rest, what is its velocity after 5 seconds?
- 10 m/s
- 20 m/s
- 5 m/s
- 2.5 m/s
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Answer: A. 10 m/s
First, find the acceleration using Newton's second law, $F = ma$. $a = \frac{F}{m} = \frac{100 \text{ N}}{50 \text{ kg}} = 2$ m/s$^2$. Since the body starts from rest ($u=0$), its velocity after time $t$ is given by $v = u + at$. $v = 0 + (2 \text{ m/s}^2 \times 5 \text{ s}) = 10$ m/s.
A body of mass 5 kg is moving with a velocity of 4 m/s. If a force of 10 N is applied for 3 seconds, what is the new momentum?
- 20 kg-m/s
- 50 kg-m/s
- 60 kg-m/s
- 80 kg-m/s
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Answer: B. 50 kg-m/s
The initial momentum of the body is $p_{initial} = mv = 5 \text{ kg} \times 4 \text{ m/s} = 20$ kg-m/s. The impulse applied by the force is $I = F imes t = 10 \text{ N} \times 3 \text{ s} = 30$ N-s. The change in momentum is equal to the impulse, so $\Delta p = 30$ kg-m/s. The new momentum is $p_{new} = p_{initial} + \Delta p = 20 \text{ kg-m/s} + 30 \text{ kg-m/s} = 50$ kg-m/s.
A car of mass 1200 kg is moving with a velocity of 15 m/s. If the brakes are applied and the car stops in 5 seconds, what is the force applied by the brakes?
- 3600 N
- -3600 N
- 7200 N
- -7200 N
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Answer: B. -3600 N
First, calculate the initial momentum $p_{initial} = mv = 1200 \text{ kg} imes 15 \text{ m/s} = 18000$ kg-m/s. The final momentum $p_{final} = mv = 1200 \text{ kg} imes 0 \text{ m/s} = 0$ kg-m/s (since the car stops). The impulse is the change in momentum: $I = p_{final} - p_{initial} = 0 - 18000 = -18000$ kg-m/s. Impulse is also equal to Force $\times$ time ($I = Ft$). So, $-18000 \text{ kg-m/s} = F \times 5 \text{ s}$. The force $F = \frac{-18000 \text{ kg-m/s}}{5 \text{ s}} = -3600$ N. The negative sign indicates that the force is acting in the opposite direction to the initial motion.