Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 7.43
In a scene from Mission: Impossible II (Paramount Pictures, 2000), Dougray Scott (mass 90 kg) and Tom Cruise (mass 80 kg) are shown jumping off motorcycles headed towards each other with different speeds until their bodies collide and stick in midair, completely stopping their horizontal motion.[*] Ignore friction, drag, and external forces, and treat this as a horizontal collision. If Dougray Scott's horizontal speed before the collision was 22 m/s (50 mph), then Tom Cruise's initial horizontal speed before the collision was:
(A) 20 m/s.
(B) 23 m/s.
(C) 25 m/s.
(D) 28 m/s.
[*] Adam Weiner, "Hollywood Physics—Collision: Impossible," popsci.com/entertainment-gaming/gallery/2007-09/hollywood-physics.
Correct answer: (C)
Negligible net external force and brief time duration for this collision makes the external impulse on this system zero, such that momentum is conserved. (Since Dougray Scott and Tom Cruise stick together in midair, this is a perfectly inelastic collision, such that kinetic energy is not conserved.)
But since total momentum is conserved, then:
0 = m1·(vf1 – v01) + m2·(vf2 – v02),
where for Dougray Scott (assumed to be initially moving to the right), m1 = 90 kg, v01 = +22 m/s, vf1 = 0; and for Tom Cruise (initially moving to the left), m2 = 80 kg, v02 = ?, vf2 = 0. So solving for the initial velocity vf2 of Tom Cruise:
0 = –m1·v01 – m2·v02,
m2·v02 = –m1·v01,
v02 = –m1·v01/m2 = –(90 kg)·(+22 m/s)/(80 kg) = –24.75 m/s,
or to two significant figures, Tom Cruise's initial speed (towards the left) was 25 m/s. (The reference link above discusses the implausibility of both surviving a collision like this in real-life.)
(Response (A) is m2·v01/m1; response (B) is v01·√(m1/m2); response (D) is v01·(m1/m2)2.)
Sections 70854, 70855, 73320
Exam code: quiz04w3Rc
(A) : 2 students
(B) : 8 students
(C) : 54 students
(D) : 1 student
Success level: 83%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.39