Question

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homework # 23 – conservation of momentum
handout problems:

1. while a small laboratory cart is moving over a table at constant velocity, a 10-kg mass is gently deposited on it. assuming that the mass had zero momentum to begin with, what happens to the velocity of the cart?
2. the diagram below shows a compressed spring between two carts initially at rest on a horizontal frictionless surface. after the string is cut, cart b moves to the right at a speed of 15 m/s. what must the speed and direction of cart a be after the string is cut?

diagram of cart a (2 kg) and cart b (1 kg) with a spring between them on a horizontal surface

1. a 1500-kg cart moving at a speed of 8.0 m/s collides with a 4500-kg cart at rest as shown in the following diagram. after they collide the two carts interlock and move together. calculate the speed of the combined carts after the collision.

diagram of 1500-kg cart moving with v_i = 8.0 m/s towards a 4500-kg cart at rest on a horizontal surface

Explanation:

Step1: Define conservation of momentum

Initial total momentum = Final total momentum. Let cart mass = $m$, initial velocity = $v_i$, final velocity = $v_f$.
Initial momentum: $m v_i + 10 \cdot 0 = m v_i$
Final momentum: $(m + 10) v_f$
Set equal: $m v_i = (m + 10) v_f$
Solve for $v_f$: $v_f = \frac{m v_i}{m + 10}$

Step2: Analyze velocity change

Since $m + 10 > m$, $v_f < v_i$.
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Step1: Set up momentum conservation

Initial total momentum = 0 (both at rest). Let cart A velocity = $v_A$, cart B velocity $v_B = 15$ m/s right.
$m_A v_A + m_B v_B = 0$
$2 v_A + 1 \cdot 15 = 0$

Step2: Solve for $v_A$

$2 v_A = -15$
$v_A = \frac{-15}{2} = -7.5$ m/s
Negative sign = direction left.
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Step1: State conservation of momentum

Initial total momentum = Final total momentum. Let combined speed = $v_f$.
Initial momentum: $m_1 v_1 + m_2 v_2 = 1500 \cdot 8.0 + 4500 \cdot 0$
Final momentum: $(m_1 + m_2) v_f = (1500 + 4500) v_f$

Step2: Calculate initial momentum

$1500 \cdot 8.0 = 12000$ kg·m/s

Step3: Solve for $v_f$

$12000 = 6000 v_f$
$v_f = \frac{12000}{6000} = 2.0$ m/s

Answer:

1. The velocity of the cart decreases, and its new velocity is given by $v_f = \frac{m v_i}{m + 10}$ where $m$ is the cart's original mass and $v_i$ is its initial velocity.
2. Cart A moves to the left at a speed of 7.5 m/s.
3. The speed of the combined carts after the collision is 2.0 m/s.