Question 5 of 6: Question 5 (paper Question V): Impact of two blocks on an incline
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3 (Statics and Dynamics) — 2016-May. 3 hours, closed book (one 8.5″×11″ formula sheet permitted). Part A (Statics) offers 3 questions, answer any 2; Part B (Dynamics) offers 3 questions, answer any 2. Every printed question is solved below (all 6), even though the exam instructions require only 4.
Given. Blocks A and B on a 20° incline, 60 in apart (A below B). Block A moves up-slope at constant velocity $v_A$; block B is released from rest and slides down-slope under gravity and friction. They collide after B has moved 20 in down-slope (so A has moved $60-20=40$ in up-slope). $\mu_k=0.12$ under A, $\mu_k=0.08$ under B. Impact is perfectly elastic ($e=1$).
Given data
Quantity
Value
Incline angle, $\theta$
20°
Weight of A, $W_A$
10 lb$_f$
Weight of B, $W_B$
12 lb$_f$
$\mu_k$ under A / under B
0.12 / 0.08
Distance B travels to impact
20 in
Distance A travels to impact
40 in
Find. (A) velocities of A and B immediately after impact; (B) the impulse delivered during the collision.
Fig. 5 – block A climbing at constant velocity, block B released from rest 60 in up-slope; they collide 20 in below B's start.
Approach. First find block B's constant deceleration-free slide from rest (kinematics with friction) to get the time-to-collision and B's speed just before impact; use that same time with A's known 40-in travel at constant speed to back out $v_A$. Then apply the 1-D elastic-collision equations (momentum + restitution $e=1$) along the incline, and finally read the impulse off the momentum change of either block.
Block B's acceleration down-slope (released from rest, friction opposes the downward slide):
$$a_B=g(\sin\theta-\mu_{k,B}\cos\theta)=386.4(\sin20^{\circ}-0.08\cos20^{\circ})=103.1\text{ in/s}^2$$
Time to collision from $d_B=\tfrac12a_Bt^2$ with $d_B=20$ in:
$$t=\sqrt{2(20)/103.1}=0.6228\text{ s}\ \Rightarrow\ v_{B,\text{before}}=a_Bt=64.22\text{ in/s}=5.352\text{ ft/s (down-slope)}$$
Block A's constant speed from $d_A=v_At$ with $d_A=40$ in:
$$v_{A,\text{before}}=40/0.6228=64.22\text{ in/s}=5.352\text{ ft/s (up-slope)}$$
(A and B happen to reach the same speed magnitude – a direct consequence of $d_A=2d_B$ in this problem.)
Elastic collision along the incline (up-slope positive; $m=W/g$): with $v_{A0}=+5.352$, $v_{B0}=-5.352$ ft/s,
$$v_A'=\frac{(m_A-m_B)v_{A0}+2m_Bv_{B0}}{m_A+m_B},\qquad v_B'=\frac{(m_B-m_A)v_{B0}+2m_Av_{A0}}{m_A+m_B}$$
$$\boxed{v_A'=-6.325\text{ ft/s}}\ (\text{i.e. 6.32 ft/s down-slope}),\qquad \boxed{v_B'=4.379\text{ ft/s (up-slope)}}$$
Impulse. Impulse on A $=m_A(v_A'-v_{A0})$, with $m_A=W_A/g=10/32.2=0.3106$ slug:
$$\boxed{\mathrm{Imp}=0.3106(-6.325-5.352)=-3.63\text{ lb}_f\cdot\text{s}}$$
(magnitude 3.63 lb$_f\cdot$s, directed down-slope on A; by Newton's third law, +3.63 lb$_f\cdot$s up-slope on B.)
