Question 4 of 6: Cart and Ball Coupled by an Elastic Cord — Impulse-Momentum in a Weightless Environment

Question 4: Cart and Ball Coupled by an Elastic Cord — Impulse-Momentum in a Weightless Environment (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. Cart (1 kg, on frictionless rollers, horizontal motion only) and ball (0.1 kg) joined by an elastic cord at the cart’s C.M. height; system starts at rest; cord exerts $\vec F_s = -1\vec\imath$ N on the cart at $t=0$; ball hits the cart’s 60°-from-vertical face at $t=2$ s and sticks.

Find. (a) whether the cart initially moves right; (b) the impulsive forces acting at the instant of impact; (c) the cart’s speed immediately after the ball sticks to it.

Approach. Part (a) follows directly from $\vec F=m\vec a$ at $t=0^+$. Parts (b)–(c) exploit that the rollers exert force only in $y$ and the cord and the impact are both internal, so the system’s TOTAL horizontal momentum is conserved from $t=0$ onward — a route to the final speed that never needs the cord’s (unstated) stiffness.

1. Part (a) — direction of initial motion. $$\vec a_{\text{cart}}(0^+) = \frac{\vec F_s}{m_{\text{cart}}} = \frac{-1\vec\imath}{1} = -1\vec\imath \text{ m/s}^2$$ $$\boxed{\text{No: the cart accelerates in the } -x \text{ direction (LEFT), not right.}}$$ By Newton’s third law the cord pulls the ball with $+1\vec\imath$ N, so the ball accelerates at $+10\ \text{m/s}^2$ toward the cart — consistent with a stretched cord pulling its two ends together (cart is to the ball’s right, so the cart is pulled left, toward the ball).
2. Part (b) — momentum bookkeeping through the impact. The rollers are frictionless and act only in $y$; the cord is an internal force pair; the impact is also internal. So no external horizontal force EVER acts on the ball+cart system, and $$m_{\text{ball}}v_{\text{ball},x}(t) + m_{\text{cart}}v_{\text{cart},x}(t) = 0 \quad \text{for every } t > 0$$ Since the cord stays horizontal throughout (ball and cart remain at the same height, the C.M. level) the ball travels in a straight horizontal line and strikes the cart’s angled face moving purely in $+x$: $\vec v_{\text{ball}}^- = (v,0)$ for some (unstated) impact speed $v$. Decomposing the ball’s impulsive contact force onto the face’s outward normal $\hat n = (-\cos60^\circ,\sin60^\circ)$ and tangent $\hat t = (\sin60^\circ,\cos60^\circ)$, and requiring the net impulse on the ball to be purely horizontal (since $v_{\text{ball},y}\equiv0$ throughout, so it starts AND ends at zero): $$J_N\hat n + J_f\hat t = (-m_{\text{ball}}v,\ 0) \;\Longrightarrow\; J_N = 0.5\,(m_{\text{ball}}v), \quad J_f = -0.866\,(m_{\text{ball}}v)$$ $$\boxed{\text{Two impulsive forces act: a NORMAL contact impulse } J_N \text{ and a TANGENTIAL (friction/adhesion) impulse } J_f \text{ needed to make the ball stick}}$$ A frictionless (normal-only) impact could not produce a purely horizontal resultant on an angled face, so “sticks to the cart surface” necessarily implies a nonzero tangential capture force here. No separate impulsive roller reaction is needed: the system’s vertical momentum is zero throughout (nothing ever pushes the ball off the horizontal line through the C.M.), so the rollers never have to absorb anything.
3. Part (c) — final speed by momentum conservation. The total horizontal momentum is always zero (system starts at rest under an internal-only force set), including immediately after the perfectly-plastic collision, when ball and cart share one common velocity $v_f$: $$(m_{\text{ball}} + m_{\text{cart}})v_f = 0$$ $$\boxed{v_f = 0}$$ Right after impact the combined cart+ball is momentarily at rest — regardless of the cord’s stiffness or the actual impact speed, both of which the source never states.