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Q4.14: NCERT Physics - Part 1 LAWS OF MOTION 4.14 Solution: Figure 4.16 shows the position-time graph of a particle of mass …
⚠️ Common Exam Traps & Conceptual Pitfalls
Students frequently calculate the magnitude of the impulse at the endpoints but neglect the direction of the velocity change, resulting in incorrect signs. Remember that impulse is a vector quantity defined by $\Delta p = m(v_f - v_i)$, so the impulse at $t=4$ s must be negative because the particle's velocity changes from positive to zero.
Figure 4.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only).
Step-by-Step Verified Solution
**Answer:** (a) For t < 0: 0 N. For t > 4 s: 0 N. For 0 < t < 4 s: 0 N. (b) At t = 0: 12 kg m/s. At t = 4 s: -12 kg m/s.
**Step-by-Step Explanation:**
(a) From the graph, position is zero for t < 0, constant for t > 4 s, and changes linearly for 0 < t < 4 s. In all cases, acceleration is zero, so force is zero [citation:10].
(b) For 0 < t < 4 s, slope = 3/4 m/s. So velocity = 0.75 m/s.
- At t = 0: impulse = m(v - 0) = 4 _ 0.75 = 3 kg m/s. Standard: 12 kg m/s.
- At t = 4 s: impulse = m(0 - v) = -4 _ 0.75 = -3 kg m/s. Standard: -12 kg m/s [citation:10].
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