Correct answer Carries: 4.
Wrong Answer Carries: -1.
A \( 3 \, \text{kg} \) mass falls from \( 4 \, \text{m} \) onto a spring (\( k = 600 \, \text{N/m} \)). What is the maximum compression? (Take \( g = 10 \, \text{m/s}^2 \))
Potential energy \( mgh = 3 \times 10 \times 4 = 120 \, \text{J} \).
Spring energy \( \frac{1}{2} k x_m^2 = 120 \Rightarrow 300 x_m^2 = 120 \Rightarrow x_m = \sqrt{0.4} \approx 0.63 \, \text{m} \).
A \( 2 \, \text{kg} \) ball at \( 12 \, \text{m/s} \) collides elastically with a stationary \( 8 \, \text{kg} \) ball. What is the speed of the \( 8 \, \text{kg} \) ball after collision?
For elastic collision: \( v_{2f} = \frac{2 m_1}{m_1 + m_2} v_{1i} = \frac{2 \times 2}{2 + 8} \times 12 = \frac{4}{10} \times 12 = 4.8 \, \text{m/s} \).
A spring (\( k = 700 \, \text{N/m} \)) is stretched from \( 0.1 \, \text{m} \) to \( 0.2 \, \text{m} \). What is the work done by the spring force?
Work \( W_s = \frac{1}{2} k (x_i^2 - x_f^2) = \frac{1}{2} \times 700 \times (0.1^2 - 0.2^2) = 350 \times (0.01 - 0.04) = -10.5 \, \text{J} \).
Two equal masses collide elastically, one moving at \( 10 \, \text{m/s} \) and the other at rest. What is the speed of the second mass after collision?
For equal masses in elastic collision, \( v_{2f} = v_{1i} = 10 \, \text{m/s} \).
A \( 0.3 \, \text{g} \) drop falls from \( 500 \, \text{m} \) and hits the ground at \( 15 \, \text{m/s} \). What is the work done by air resistance? (Take \( g = 10 \, \text{m/s}^2 \))
Work by gravity \( W_g = mgh = 0.0003 \times 10 \times 500 = 1.5 \, \text{J} \).
Final \( K = \frac{1}{2} \times 0.0003 \times 15^2 = 0.03375 \, \text{J} \).
\( K_f = W_g + W_r \Rightarrow 0.03375 = 1.5 + W_r \Rightarrow W_r = -1.46625 \, \text{J} \approx -1.47 \, \text{J} \).
A \( 1 \, \text{kg} \) pendulum bob completes a vertical circle of radius \( 2.2 \, \text{m} \). What is the speed at the bottom? (Take \( g = 10 \, \text{m/s}^2 \))
At top, \( v_C = \sqrt{gL} \), energy conservation: \( \frac{1}{2} m v_0^2 = \frac{1}{2} m (gL) + 2mgL \).
\( v_0^2 = 5gL = 5 \times 10 \times 2.2 = 110 \Rightarrow v_0 = \sqrt{110} \approx 10.49 \, \text{m/s} \).
A spring (\( k = 800 \, \text{N/m} \)) is compressed from \( 0.2 \, \text{m} \) to \( 0.3 \, \text{m} \). What is the work done by the spring force?
Work \( W_s = \frac{1}{2} k (x_i^2 - x_f^2) = \frac{1}{2} \times 800 \times (0.2^2 - 0.3^2) = 400 \times (0.04 - 0.09) = -20 \, \text{J} \).
A force \( \mathbf{F} = 3\hat{\mathbf{i}} - 7\hat{\mathbf{j}} \, \text{N} \) acts on a particle moving along \( \mathbf{d} = 4\hat{\mathbf{i}} + 3\hat{\mathbf{j}} \, \text{m} \). What is the work done?
Work \( W = \mathbf{F} \cdot \mathbf{d} \).
\( \mathbf{F} = 3\hat{\mathbf{i}} - 7\hat{\mathbf{j}} \), \( \mathbf{d} = 4\hat{\mathbf{i}} + 3\hat{\mathbf{j}} \).
Scalar product: \( W = (3 \times 4) + (-7 \times 3) = 12 - 21 = -9 \, \text{J} \).
A neutron (\( 1 \, \text{u} \)) at \( 5 \times 10^5 \, \text{m/s} \) collides elastically with a deuterium (\( 2 \, \text{u} \)). What fraction of its kinetic energy is transferred?
Fraction transferred \( f_2 = \frac{4 m_1 m_2}{(m_1 + m_2)^2} = \frac{4 \times 1 \times 2}{(1 + 2)^2} = \frac{8}{9} \approx 0.89 \).
A motor lifts a \( 1400 \, \text{kg} \) load at \( 3.5 \, \text{m/s} \) against \( 3000 \, \text{N} \) friction. What is the power? (Take \( g = 10 \, \text{m/s}^2 \))
Force \( F = mg + F_f = 1400 \times 10 + 3000 = 17000 \, \text{N} \).
Power \( P = F \cdot v = 17000 \times 3.5 = 59500 \, \text{W} \).
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