The kinetic energy of a body is 100 J and momentum 50 kg m/s. Mass is MCQ with Answer and Explanation

Explanation:
Position-time graph as parabola: x = x₀ + ut + ½at². This quadratic form indicates constant acceleration 'a'. If acceleration were changing, the graph would be cubic or higher order. Constant velocity would produce a straight line (linear graph). Upward opening parabola implies positive acceleration. Graph interpretation skill is essential: slope of x-t graph = velocity; curvature indicates acceleration. Memory tip: Parabolic x-t graph ⇔ constant acceleration; linear x-t graph ⇔ constant velocity. Competitive exams frequently test graph-concept correlations.

Explanation:
Electric power P = VI. Thus I = P/V = 100 W / 220 V ≈ 0.4545 A ≈ 0.45 A. This direct application of power formula is fundamental in electricity. Memory tip: 'I = P/V for resistive loads'. Competitive exams frequently test such calculations with household appliance ratings. Always use consistent units (watts, volts, amperes). Note: This assumes purely resistive load (valid for incandescent bulbs); for motors or electronics, power factor may matter, but not at this level.

Explanation:
By conservation of momentum: initial momentum = 0 (system at rest). Final momentum: m_bullet×v_bullet + m_gun×v_gun = 0. Thus (0.02 kg)(300 m/s) + (2 kg)(v_gun) = 0 ⇒ 6 + 2v_gun = 0 ⇒ v_gun = -3 m/s. Magnitude is 3 m/s; negative sign indicates opposite direction to bullet. This demonstrates momentum conservation in isolated systems. Memory tip: Recoil velocity = -(m_bullet/m_gun)×v_bullet. Such numerical problems test application of conservation laws, frequently appearing in competitive exams with varying mass ratios and velocities.