Question

a 7.0-kilogram cart, a, and a 3.0-kilogram cart, b, are initially held together at rest on a horizontal, frictionless surface. when a compressed spring attached to one of the carts is released, the carts are pushed apart. after the spring is released, the speed of cart b is 6.0 meters per second, as represented in the diagram below.what is the speed of cart a after the spring is released?a. 14 m/sb. 6.0 m/sc. 3.0 m/sd. 2.6 m/s

Explanation:

Step1: Apply conservation of momentum

Since the system is initially at rest, total initial momentum is 0. For a closed, frictionless system, total final momentum equals initial momentum: $m_A v_A + m_B v_B = 0$

Step2: Rearrange to solve for $v_A$

Isolate $v_A$: $v_A = -\frac{m_B v_B}{m_A}$

Step3: Substitute given values

$m_A=7.0\ \text{kg}$, $m_B=3.0\ \text{kg}$, $v_B=6.0\ \text{m/s}$
$v_A = -\frac{3.0 \times 6.0}{7.0} = -\frac{18}{7.0} \approx -2.6\ \text{m/s}$
The negative sign indicates direction opposite to cart B; speed is the magnitude.

Answer:

D. 2.6 m/s