Question 4 of 6: Ball, Elastic Cord and Cart in a Weightless Environment — Impulse and Momentum

Question 4: Ball, Elastic Cord and Cart in a Weightless Environment — Impulse and Momentum (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Ball $m_{\text{ball}}=0.1$ kg, cart $m_{\text{cart}}=1$ kg, connected by an elastic cord attached at the cart's C.M. height (so the cord, and hence its tension, stays horizontal); no gravity; the cart's rollers permit only horizontal cart motion (zero net horizontal external force is possible from them); at $t=0$ both bodies are at rest with the cord exerting $\vec F_s=-1\vec i$ N on the cart; the ball strikes and sticks to the cart at $t=2$ s.

Find. (a) direction of the cart's initial motion; (b) the impulsive force(s) at impact; (c) the common speed of ball+cart immediately after the sticking collision.

Approach. The cord tension and, later, the sticking-contact force are both internal to the (ball + cart) system; the rollers can only supply a force perpendicular to the direction of cart travel. With no gravity and no other external agent, the total horizontal momentum of the (ball + cart) system is therefore conserved from release all the way through the collision — and since both bodies start at rest, that total is zero for all time, including immediately after they stick together.

1. Part (a). At $t=0$ the cord is taut and pulls the cart with $\vec F_s=-1\vec i$ N (toward the ball). An elastic cord's tension depends on its stretch, and the stretch cannot change in zero time, so the force an instant later ($t=0^+$) is still $-1\vec i$ N (its magnitude only begins to decrease as the cord shortens). Once the ball is released the cart is no longer held in equilibrium, so Newton's second law gives $\vec a_{\text{cart}}(0^+)=\vec F_s/m_{\text{cart}}=-1\vec i$ m/s². Yes — the cart starts to move immediately at $t=0^+$, accelerating at 1 m/s² to the left ($-\vec i$), toward the ball, because the stretched elastic cord always pulls its two ends toward one another (contrast an inextensible string, whose tension could drop to zero instantaneously).
2. Part (b). The only forces available during the brief sticking collision are (i) an internal, equal-and-opposite impulsive contact/sticking force between the ball and the cart (Newton's third law pair, directed along the line of impact, i.e.\ horizontally, since the ball's flight stays on the cord's original horizontal line by the C.M.-height condition given), and (ii) the roller reactions, which stay purely perpendicular to the direction of travel and therefore carry zero impulsive component along the direction of motion (there is nothing to arrest in that direction — the whole collision is 1-D horizontal). On the slanted face (60° from vertical) the contact impulse has a component normal to the face and a tangential (sticking) component; because both bodies move only horizontally before and after the impact, these combine to a purely horizontal resultant, so no vertical impulsive roller reaction is needed. So: a single internal horizontal impulsive pair between ball and cart; no impulsive force from the rollers.
3. Part (c) — momentum conservation through the whole process. From $t=0$ (both at rest, $\Sigma p_x=0$) to just before impact at $t=2$ s, only the internal cord tension acts on the two-body system, so $\Sigma p_x$ remains exactly $0$ the entire time: $$m_{\text{ball}}v_{\text{ball}} + m_{\text{cart}}v_{\text{cart}} = 0 \quad\text{(true at every instant, including }t=2^-\text{s).}$$ The sticking collision itself is likewise governed only by the same internal pair, so total momentum is unchanged through it as well. Immediately after impact the two move together at a common speed $v$: $$(m_{\text{ball}}+m_{\text{cart}})\,v = m_{\text{ball}}v_{\text{ball}}+m_{\text{cart}}v_{\text{cart}} = 0 \quad\Longrightarrow\quad \boxed{v=0}.$$ This holds regardless of the (unknown, and unneeded) individual pre-impact speeds or the cord's spring constant — the whole ball+cart system was never given any net momentum, so it cannot have any immediately after they lock together.