Question

80 kg adventurer ties a rope of length L to a bridgs and falls off in s wing figure. The tension (in N) in the rope at the lowest point is ewn in the a) 2100 b) 2200 c) 2300 d) 2400 e) 2500 Block A with mass ma 2 kg traveling at u 10 mvs to the right collides with block B with mass mp 3 kg traveling at Un 5 m/s to the left. Afier the colision block A travels with a speed of A m/s to the left and block B with a speed of 's 7m's to the right. Which of the following staiemeni is correct? mtB a) The collision is a perfectly inclastic collision. b) The collision is an inelastic collision but not perfectly inelastic. c) The collision is an elastic collision. d) The collision is not possible. c) None of the above. A B block of mass m 2 kg is held compressing a spring of force constant k - 800 0.1 distance through the I m on a rough surface as shown in the figure. The block is then released and goes cal loop of radius r-0.5 m. If the speed of the block at the top of the loop (point B) then how many joules are lost due to friction? is u b) 21 c) 19 d) 15 e) 11 A force-time curve for a baseball of mass 5 g struck by a bat is shown in the figure itude of the change in its momentum (in N.s)? FIN) a) 50 b) 100 c) 150 d) 200 e) 250 23

Answer 1

Tension = mv^2/L + mg m = 80 kg, g = 10 m/s^2, Conservation of energy, mg L = 1/2 mv^2 mv^2/L = 2mg T = 2 mg + mg = 3 mg = 3 times 10 times 80 = 2400 N option (d) is correct. m_A v_A + m_B v_B = 2 times 10 + 3(-5) = 5 kg m/s m_A v_A' + m_B v_B' = 2 times (-8) + 3(7) = 5 kg m/s Momentum is conserved, collision is possible Initial KE = 1/2 times 2 times 100 + 1/2 times 3 times 25 = 137.5 J Final KE = 1/2 times 2 times 64 + 1/2 times 3 times 49 = 137.5J KE is conserved collision is elastic (c) collision is an elastic collision. 1/2 kx^2 = mg (2r) + 1/2 mv^2 + w_f 1/2 times 8000 times 0.1^2 = 2 times 10 times 2 times 0.5 + 1/2 times 2 times 9 + w_f