A student investigates the relationship between mass and recoil speed using a spring-loaded cart. The cart sits on a smooth horizontal surface and has a com...

(a) Recoil speed of the cart in trial 1 [2]

Before release, both the cart and block are stationary, so total momentum = 0. By conservation of momentum, total momentum after release must also be zero:

\(0 = m_{\text{block}} v_{\text{block}} + m_{\text{cart}} v_{\text{cart}}\)

\(0 = (0.20 \times 2.0) + (0.80 \times v_{\text{cart}})\)

\(0 = 0.40 + 0.80 v_{\text{cart}}\)

\(v_{\text{cart}} = \dfrac{-0.40}{0.80} = -0.50\) m/s

The negative sign means the cart moves to the left (opposite to the block). The cart recoils at 0.50 m/s. The cart is four times heavier than the block, so it moves at one-quarter of the block's speed.

(b) Showing consistency with 0.50 J of spring energy [2]

The total kinetic energy after release should equal the elastic potential energy stored in the spring:

\(KE_{\text{total}} = \frac{1}{2} \times 0.20 \times 2.0^2 + \frac{1}{2} \times 0.80 \times 0.50^2\)

\(= 0.40 + 0.10 = 0.50\) J

This equals the spring's stored energy of 0.50 J, confirming consistency. All the elastic potential energy has been converted to kinetic energy (no losses on the frictionless surface).

(c) Trial 2: block mass doubled to 0.40 kg [4]

The spring compression is the same, so total KE = 0.50 J. Two equations are needed (momentum conservation and energy conservation).

Momentum conservation (total momentum = 0):

\(0 = 0.40 v_b + 0.80 v_c\)

\(v_c = -0.50 v_b\)

Energy conservation:

\(\frac{1}{2} \times 0.40 \times v_b^2 + \frac{1}{2} \times 0.80 \times v_c^2 = 0.50\)

Substituting \(v_c = -0.50 v_b\):

\(0.20 v_b^2 + 0.40 \times (0.50 v_b)^2 = 0.50\)

\(0.20 v_b^2 + 0.40 \times 0.25 v_b^2 = 0.50\)

\(0.20 v_b^2 + 0.10 v_b^2 = 0.50\)

\(0.30 v_b^2 = 0.50\)

\(v_b^2 = 1.667\)

\(v_b = 1.29\) m/s (accept 1.3 m/s) - block speed to the right

\(v_c = -0.50 \times 1.29 = -0.645\) m/s, so cart recoils at 0.65 m/s to the left

Doubling the block mass reduces the block's launch speed (from 2.0 to 1.3 m/s) but increases the cart's recoil speed (from 0.50 to 0.65 m/s). The heavier block takes more of the spring's energy as its own KE, leaving less for the cart, but the momentum constraint forces the cart to move faster to compensate for the heavier block's momentum.