Newton's First Law — Question 9

Question 9

Releasing a moving object preserves its velocity at release

A cart moves at 5 m/s on an ideal level track. It releases a small slider with the same horizontal velocity. After release the slider has no net horizontal force; vertical forces balance. Set both horizontal positions to zero at release.

1. Find slider position after 3 s.

2. If the cart brakes uniformly at 1 m/s$^2$, find their separation after 3 s.

3. Why does the slider move forward relative to the braking cart?

Question 9 - Solution

Releasing a moving object preserves its velocity at release

(a) Preserve the slider’s velocity.

$$x_S(t)=v_0t,\quad\boxed{x_S(3)=5(3)=15\,\mathrm m}$$ The slider follows $x_S=5t$ m, hence reaches 15 m in 3 s.

(b) Evolve the cart separately.

$$x_C=5t-\tfrac12t^2,\quad x_C(3)=10.5\,\mathrm m,\quad x_S-x_C=4.5\,\mathrm m$$ $x_C=5t-\tfrac12t^2$ m, so $x_C(3)=10.5$ m. The slider is 4.5 m ahead.

(c) Attribute the relative change correctly.

$$a_{S/C}=a_S-a_C=0-(-1)=1\,\mathrm{m/s^2},\quad F_{S,x}=0$$ The cart’s velocity changes while the slider’s remains constant. A forward slider motion relative to the cart does not require a new forward force on the slider; the observer is decelerating.