Calculate the speed of the 1.5 kg trolley at point P using the principle of conservation of mechanical energy.

Question

Calculate the speed of the 1.5 kg trolley at point P using the principle of conservation of mechanical energy.

Asked on March 28, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-03-28T12:57:36.195Z

Step 1: Calculate the speed of the 1.5 kg trolley at point P using the principle of conservation of mechanical energy. Given: Mass of trolley 1, $m_1 = 1.5$ kg Initial height at A, $h_A = 2.0$ m Initial velocity at A, $v_A = 0$ m/s (held stationary) Height at P, $h_P = 0$ m (bottom of the incline) Acceleration due to gravity, $g = 9.8$ m/s$^2$ According to the principle of conservation of mechanical energy: $$PE_A + KE_A = PE_P + KE_P$$ $$m_1 g h_A + \frac{1}{2} m_1 v_A^2 = m_1 g h_P + \frac{1}{2} m_1 v_P^2$$ Substitute the known values: $$(1.5 \text{ kg})(9.8 \text{ m/s}^2)(2.0 \text{ m}) + \frac{1}{2} (1.5 \text{ kg})(0 \text{ m/s})^2 = (1.5 \text{ kg})(9.8 \text{ m/s}^2)(0 \text{ m}) + \frac{1}{2} (1.5 \text{ kg}) v_P^2$$ $$29.4 \text{ J} + 0 = 0 + 0.75 v_P^2$$ $$29.4 = 0.75 v_P^2$$ $$v_P^2 = \frac{29.4}{0.75}$$ $$v_P^2 = 39.2$$ $$v_P = \sqrt{39.2}$$ $$v_P \approx 6.26 \text{ m/s}$$ The speed of the 1.5 kg trolley at point P is $\boxed{\text{6.26 m/s}}$. Step 2: Rewrite the complete statement for the principle of conservation of linear momentum. In an isolated system, the total linear momentum is conserved. Step 3: Calculate the speed of the combined trolleys immediately after the collision. This is an inelastic collision where the trolleys stick together. Given: Mass of trolley 1, $m_1 = 1.5$ kg Velocity of trolley 1 before collision, $v_{1i} = v_P = 6.26$ m/s (from Step 1) Mass of trolley 2, $m_2 = 2$ kg Velocity of trolley 2 before collision, $v_{2i} = 0$ m/s (stationary) Combined mass after collision, $M = m_1 + m_2 = 1.5 \text{ kg} + 2 \text{ kg} = 3.5$ kg Let $v_f$ be the final velocity of the combined trolleys. According to the principle of conservation of linear momentum: $$m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2) v_f$$ Substitute the known values: $$(1.5 \text{ kg})(6.26 \text{ m/s}) + (2 \text{ kg})(0 \text{ m/s}) = (1.5 \text{ kg} + 2 \text{ kg}) v_f$$ $$9.39 \text{ kg}\cdot\text{m/s} + 0 = (3.5 \text{ kg}) v_f$$ $$9.39 = 3.5 v_f$$ $$v_f = \frac{9.39}{3.5}$$ $$v_f \approx 2.68 \text{ m/s}$$ The speed of the combined trolleys immediately after the collision is $\boxed{\text{2.68 m/s}}$. Step 4: Calculate the distance travelled by the combined trolleys in 3 s after the collision. After the collision, the combined trolleys move with constant velocity. Given: Velocity of combined trolleys, $v_f = 2.68$ m/s (from Step 3) Time, $t = 3$ s Distance travelled, $d = v_f \times t$ $$d = (2.68 \text{ m/s})(3 \text{ s})$$ $$d = 8.04 \text{ m}$$ The distance travelled by the combined trolleys in 3 s after the collision is $\boxed{\text{8.04 m}}$.

Accepted Answer · 2026-03-28T12:57:36.195Z

Step 1: Calculate the speed of the 1.5 kg trolley at point P using the principle of conservation of mechanical energy. Given: Mass of trolley 1, $m_1 = 1.5$ kg Initial height at A, $h_A = 2.0$ m Initial velocity at A, $v_A = 0$ m/s (held stationary) Height at P, $h_P = 0$ m (bottom of the incline) Acceleration due to gravity, $g = 9.8$ m/s$^2$ According to the principle of conservation of mechanical energy: $$PE_A + KE_A = PE_P + KE_P$$ $$m_1 g h_A + \frac{1}{2} m_1 v_A^2 = m_1 g h_P + \frac{1}{2} m_1 v_P^2$$ Substitute the known values: $$(1.5 \text{ kg})(9.8 \text{ m/s}^2)(2.0 \text{ m}) + \frac{1}{2} (1.5 \text{ kg})(0 \text{ m/s})^2 = (1.5 \text{ kg})(9.8 \text{ m/s}^2)(0 \text{ m}) + \frac{1}{2} (1.5 \text{ kg}) v_P^2$$ $$29.4 \text{ J} + 0 = 0 + 0.75 v_P^2$$ $$29.4 = 0.75 v_P^2$$ $$v_P^2 = \frac{29.4}{0.75}$$ $$v_P^2 = 39.2$$ $$v_P = \sqrt{39.2}$$ $$v_P \approx 6.26 \text{ m/s}$$ The speed of the 1.5 kg trolley at point P is $\boxed{\text{6.26 m/s}}$. Step 2: Rewrite the complete statement for the principle of conservation of linear momentum. In an isolated system, the total linear momentum is conserved. Step 3: Calculate the speed of the combined trolleys immediately after the collision. This is an inelastic collision where the trolleys stick together. Given: Mass of trolley 1, $m_1 = 1.5$ kg Velocity of trolley 1 before collision, $v_{1i} = v_P = 6.26$ m/s (from Step 1) Mass of trolley 2, $m_2 = 2$ kg Velocity of trolley 2 before collision, $v_{2i} = 0$ m/s (stationary) Combined mass after collision, $M = m_1 + m_2 = 1.5 \text{ kg} + 2 \text{ kg} = 3.5$ kg Let $v_f$ be the final velocity of the combined trolleys. According to the principle of conservation of linear momentum: $$m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2) v_f$$ Substitute the known values: $$(1.5 \text{ kg})(6.26 \text{ m/s}) + (2 \text{ kg})(0 \text{ m/s}) = (1.5 \text{ kg} + 2 \text{ kg}) v_f$$ $$9.39 \text{ kg}\cdot\text{m/s} + 0 = (3.5 \text{ kg}) v_f$$ $$9.39 = 3.5 v_f$$ $$v_f = \frac{9.39}{3.5}$$ $$v_f \approx 2.68 \text{ m/s}$$ The speed of the combined trolleys immediately after the collision is $\boxed{\text{2.68 m/s}}$. Step 4: Calculate the distance travelled by the combined trolleys in 3 s after the collision. After the collision, the combined trolleys move with constant velocity. Given: Velocity of combined trolleys, $v_f = 2.68$ m/s (from Step 3) Time, $t = 3$ s Distance travelled, $d = v_f \times t$ $$d = (2.68 \text{ m/s})(3 \text{ s})$$ $$d = 8.04 \text{ m}$$ The distance travelled by the combined trolleys in 3 s after the collision is $\boxed{\text{8.04 m}}$.