Question Detail

PHY 101 PHY 101 Past Questions Objective Question
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Question

A 2500 kg body travelling at 40 km/h strikes a concrete wall and comes to rest in 0.2 s. What is the change in momentum?
Options
A 1.78 × 10⁴ kg m/s
B 2.78 × 10⁴ kg m/s
Correct Answer
C 2.58 × 10⁴ kg m/s
D 0.78 × 10⁴ kg m/s
Correct Answer

Option B is the correct answer.

Detailed Explanation

Convert 40 km/h to m/s: 40 × (1000/3600) = 40 × 0.2778 = 11.11 m/s. Initial momentum = 2500 × 11.11 = 27775 kg m/s ≈ 2.78 × 10⁴ kg m/s. Final momentum = 0. Change in momentum = 2.78 × 10⁴ kg m/s.

Hint

Δp = m(v_f - v_i).

Question Info

Course: PHY 101
Topic: PHY 101 Past Questions
Type: Objective
Difficulty: Not specified
Question 948 of 1250
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