Question:
A car of mass 1800 kg travelling north at 30 km h\(^{-1}\) turns east and accelerates to 40 km h\(^{-1}\). The magnitude of the change in its momentum is in kg m s\(^{-1}\).
A car of mass 1800 kg travelling north at 30 km h\(^{-1}\) turns east and accelerates to 40 km h\(^{-1}\). The magnitude of the change in its momentum is in kg m s\(^{-1}\).
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Whenever direction changes, treat momentum as vectors and use Pythagoras theorem instead of simple subtraction.
Updated On: Jun 20, 2026
- 5000
- 25000
- 10000
- 20000
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The Correct Option is B
Solution and Explanation
Step 1: Understand the concept.
Momentum is a vector quantity. So when direction changes, we must use vector subtraction: \[ \Delta \vec{p} = \vec{p_f} - \vec{p_i} \] Magnitude is found using Pythagoras theorem.
Step 2: Convert velocities into SI units.
\[ 30 \, \text{km h}^{-1} = 30 \times \frac{5}{18} = \frac{150}{18} = 8.33 \, \text{m s}^{-1} \] \[ 40 \, \text{km h}^{-1} = 40 \times \frac{5}{18} = \frac{200}{18} = 11.11 \, \text{m s}^{-1} \]
Step 3: Represent momentum vectors.
Take north as \(+\hat{j}\), east as \(+\hat{i}\). Initial momentum: \[ \vec{p_i} = 1800 \times 8.33 \hat{j} = 14994 \hat{j} \] Final momentum: \[ \vec{p_f} = 1800 \times 11.11 \hat{i} = 19998 \hat{i} \]
Step 4: Compute change in momentum vector.
\[ \Delta \vec{p} = 19998 \hat{i} - 14994 \hat{j} \]
Step 5: Find magnitude of change in momentum.
\[ |\Delta \vec{p}| = \sqrt{(19998)^2 + (14994)^2} \] Factor 2 significant approximation: \[ \approx \sqrt{(20000)^2 + (15000)^2} \] \[ = \sqrt{4 \times 10^8 + 2.25 \times 10^8} \] \[ = \sqrt{6.25 \times 10^8} \] \[ = 25000 \]
Step 6: Final conclusion.
Thus, magnitude of change in momentum is: \[ \boxed{25000 \, \text{kg m s}^{-1}} \]
Momentum is a vector quantity. So when direction changes, we must use vector subtraction: \[ \Delta \vec{p} = \vec{p_f} - \vec{p_i} \] Magnitude is found using Pythagoras theorem.
Step 2: Convert velocities into SI units.
\[ 30 \, \text{km h}^{-1} = 30 \times \frac{5}{18} = \frac{150}{18} = 8.33 \, \text{m s}^{-1} \] \[ 40 \, \text{km h}^{-1} = 40 \times \frac{5}{18} = \frac{200}{18} = 11.11 \, \text{m s}^{-1} \]
Step 3: Represent momentum vectors.
Take north as \(+\hat{j}\), east as \(+\hat{i}\). Initial momentum: \[ \vec{p_i} = 1800 \times 8.33 \hat{j} = 14994 \hat{j} \] Final momentum: \[ \vec{p_f} = 1800 \times 11.11 \hat{i} = 19998 \hat{i} \]
Step 4: Compute change in momentum vector.
\[ \Delta \vec{p} = 19998 \hat{i} - 14994 \hat{j} \]
Step 5: Find magnitude of change in momentum.
\[ |\Delta \vec{p}| = \sqrt{(19998)^2 + (14994)^2} \] Factor 2 significant approximation: \[ \approx \sqrt{(20000)^2 + (15000)^2} \] \[ = \sqrt{4 \times 10^8 + 2.25 \times 10^8} \] \[ = \sqrt{6.25 \times 10^8} \] \[ = 25000 \]
Step 6: Final conclusion.
Thus, magnitude of change in momentum is: \[ \boxed{25000 \, \text{kg m s}^{-1}} \]
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