University:
California State University, Northridge
Course:
PHYS 100A | General Physics I
Academic year:

2016
Views:

456

Pages:

6
Author:

shushifiedewv

Header: Physics 100A - S16 - Midterm 3 A 07/05/16 Instructions: In order to get full credit for your work please show all your work, draw necessary diagrams and explain reasoning clearly where necessary. Problem 1: On a very muddy football field, a 110-kg linebacker tackles an 80-kg halfback. Immediately before the collision, the linebacker is slipping with a velocity of 8.8 m/s north and the halfback is sliding with a velocity of 7.2 m/s west. What is the velocity (magnitude and direction) at which the two players move together immediately after the collision? (5.0 points) Given: m_L = 110 kg m_H = 80 kg v_L = 8.8 m/s north v_H = -7.2 m/s west Calculations: p_L = m_L * v_L p_H = m_H * v_H p_L+H = p_L + p_H (m_L + m_H) * v_fx = m_H * v_H v_fx = -3.0 m/s (m_L + m_H) * v_fy = m_L * v_L v_fy = 5.1 m/s v_f = sqrt(v_fx^2 + v_fy^2) = 5.9 m/s theta = tan^-1(v_fx / v_fy) = 30.8 degrees Header: Physics 100A - S16 - Midterm 3 A 07/05/16 Instructions: In order to get full credit for your work please show all your work, draw necessary diagrams and explain reasoning clearly where necessary. Problem 2: A basketball with a radius of 15 cm rolls on a level surface with a constant angular speed of 10 rad/s. To what height on a 30-degree inclined plane will the basketball roll before coming to rest? (I = 2/3 mr²) (5.0 points) Given: r = 15 cm = 0.15 m ω_i = 10 rad/s ω_f = 0 y_i = 0 y_f = h Calculations: K_ri + K_ki + U_i = K_rf + K_kf + U_f 1/2 mv^2 + 1/2 Iω^2 = mgh v = rω 1/2 mr^2ω^2 + 1/2 (2/3 mr^2)ω^2 = mgh r^2ω^2 (1/2 + 1/3) = gh h = (5/6) * (r^2ω^2 / g) h = (5/6) * (0.15^2 * 10^2 / 9.81) = 0.19 m Header: Physics 100A - S16 - Midterm 3 A 07/05/16 Instructions: In order to get full credit for your work please show all your work, draw necessary diagrams and explain reasoning clearly where necessary. Problem 3: A bicycle coasts downhill and accelerates from rest to a linear speed of 8.90 m/s in 12.2 s. If the bicycle's tires have a radius of 36.0 cm, what is their angular displacement during this period of acceleration? (5.0 points) Given: v_0 = 0 v = 8.90 m/s t = 12.2 s r = 36.0 cm = 0.36 m Calculations: ω = v / r = 8.90 / 0.36 = 24.7 rad/s Δθ = (ω + ω_0)t / 2 = (24.7 + 0) * 12.2 / 2 = 151 rad or 24.0 rev