A 60 kg boy throws a 5 kg package eastward with a velocity of 4 m/s. Describe the motion of the boy and the law that governs it, and calculate the boy's recoil velocity.

$$m_b u_b + m_p u_p = m_b v_b + m_p v_p$$ $$(60 \text{ kg})(0 \text{ m/s}) + (5 \text{ kg})(0 \text{ m/s}) = (60 \text{ kg}) v_b + (5 \text{ kg})(4 \text{ m/s})$$ $$0 + 0 = 60 v_b + 20$$ $$0 = 60 v_b + 20$$ $$60 v_b = -20$$ $$v_b = \frac{-20}{60}$$ $$v_b = -\frac{1}{3} \text{ m/s}$$

The magnitude of the velocity of the boy is $\boxed{\text{}0.33 \text{ m/s}\text{}}$ (rounded to two decimal places).

4.4
4.4.1 If the boy throws the same package at a higher velocity in the same direction, the momentum of the package ($m_p v_p$) will increase. To conserve the total momentum of the system (which remains zero), the magnitude of the boy's momentum ($m_b v_b$) must also increase. Since the boy's mass ($m_b$) is constant, his velocity ($v_b$) must increase in magnitude.
The answer to QUESTION 4.3 will $\boxed{\text{}INCREASES\text{}}$

4.4.2 If the boy throws a package of double the mass at the same velocity as in QUESTION 4.3, the momentum of the package ($m_p v_p$) will double. To conserve the total momentum of the system, the magnitude of the boy's momentum ($m_b v_b$) must also double. Since the boy's mass ($m_b$) is constant, his velocity ($v_b$) must double in magnitude.
The answer to QUESTION 4.3 will $\boxed{\text{}INCREASES\text{}}$
Explanation: The momentum of the package ($m_p v_p$) will double. By the principle of conservation of linear momentum, the magnitude of the boy's momentum ($m_b v_b$) must also double. Since the boy's mass ($m_b$) remains constant, his velocity ($v_b$) must increase.

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2015 Metro South Prelim Paper Q4

4.1 The principle of conservation of linear momentum states that:
$\boxed{\text{}The total linear momentum of an isolated system remains constant in both magnitude and direction.\text{}}$

4.2 Step 1: Identify the given values.
Mass of wooden block $m_w = 2 \text{ kg}$
Initial velocity of wooden block $u_w = 5 \text{ m/s}$ (to the right, let's take right as positive)
Mass of crate $m_c = 9 \text{ kg}$
Initial velocity of crate $u_c = 0 \text{ m/s}$
Final velocity of crate $v_c = 1 \text{ m/s}$ (to the right)
Final velocity of wooden block $v_w = ?$

$$m_w u_w + m_c u_c = m_w v_w + m_c v_c$$ $$(2 \text{ kg})(5 \text{ m/s}) + (9 \text{ kg})(0 \text{ m/s}) = (2 \text{ kg}) v_w + (9 \text{ kg})(1 \text{ m/s})$$ $$10 + 0 = 2 v_w + 9$$ $$10 = 2 v_w + 9$$ $$2 v_w = 10 - 9$$ $$2 v_w = 1$$ $$v_w = \frac{1}{2} \text{ m/s}$$ $$v_w = 0.5 \text{ m/s}$$

The magnitude of the velocity of the wooden block immediately after the collision is $\boxed{\text{}0.5 \text{ m/s}\text{}}$

4.3 Step 1: Identify the given values for the crate.
Mass of crate $m_c = 9 \text{ kg}$
Initial velocity of crate $u_c = 0 \text{ m/s}$
Final velocity of crate $v_c = 1 \text{ m/s}$
Time of collision $\Delta t = 0.6 \text{ s}$
Force exerted on the crate $F = ?$

$$F \Delta t = \Delta p_c$$ $$F \Delta t = m_c v_c - m_c u_c$$ $$F (0.6 \text{ s}) = (9 \text{ kg})(1 \text{ m/s}) - (9 \text{ kg})(0 \text{ m/s})$$ $$F (0.6) = 9 - 0$$ $$F (0.6) = 9$$ $$F = \frac{9}{0.6}$$ $$F = 15 \text{ N}$$

The force the wooden block exerts on the crate during the collision is $\boxed{\text{}15 \text{ N}\text{}}$