Askiitians Tutor Team
Last Activity: 9 Months ago
Let's break down the situation and solve the problem step by step.
We are given:
• Two toy carts with masses m1m_1 and m2m_2,
• A spring that is compressed between the two carts,
• When released, the spring exerts equal and opposite forces on the carts for the same time tt,
• The coefficient of friction μ\mu between the ground and the carts is equal.
We need to find the ratio of the displacements of the two toy carts, S1S2\frac{S_1}{S_2}, where S1S_1 and S2S_2 are the displacements of carts with masses m1m_1 and m2m_2, respectively.
Step 1: Force exerted by the spring on the carts
From Newton's third law, we know that the spring exerts equal and opposite forces on the two carts. Let the force exerted on cart 1 be F1F_1 and the force exerted on cart 2 be F2F_2. Since these forces are equal in magnitude and opposite in direction, we have:
F1=F2F_1 = F_2
Step 2: Frictional force on each cart
The frictional force on each cart is given by:
Ffriction=μmgF_{\text{friction}} = \mu m g
where μ\mu is the coefficient of friction, mm is the mass of the cart, and gg is the acceleration due to gravity. Since the frictional forces on the two carts are equal (because the coefficient of friction μ\mu is the same for both carts and gg is constant), the frictional forces are proportional to the masses of the carts:
Ffriction,1=μm1gandFfriction,2=μm2gF_{\text{friction,1}} = \mu m_1 g \quad \text{and} \quad F_{\text{friction,2}} = \mu m_2 g
Step 3: Equation of motion for each cart
Using Newton's second law, the acceleration of each cart is given by:
a1=F1−Ffriction,1m1=F1−μm1gm1a_1 = \frac{F_1 - F_{\text{friction,1}}}{m_1} = \frac{F_1 - \mu m_1 g}{m_1} a2=F2−Ffriction,2m2=F2−μm2gm2a_2 = \frac{F_2 - F_{\text{friction,2}}}{m_2} = \frac{F_2 - \mu m_2 g}{m_2}
Since F1=F2F_1 = F_2, the accelerations are:
a1=F1m1−μga_1 = \frac{F_1}{m_1} - \mu g a2=F2m2−μga_2 = \frac{F_2}{m_2} - \mu g
Step 4: Displacements of the carts
The displacements of the carts depend on the accelerations and the time tt they experience the forces. Using the kinematic equation for displacement, assuming the initial velocity is zero:
S1=12a1t2=12(F1m1−μg)t2S_1 = \frac{1}{2} a_1 t^2 = \frac{1}{2} \left( \frac{F_1}{m_1} - \mu g \right) t^2 S2=12a2t2=12(F2m2−μg)t2S_2 = \frac{1}{2} a_2 t^2 = \frac{1}{2} \left( \frac{F_2}{m_2} - \mu g \right) t^2
Step 5: Ratio of displacements
Now, we can find the ratio S1S2\frac{S_1}{S_2}:
S1S2=12(F1m1−μg)t212(F2m2−μg)t2\frac{S_1}{S_2} = \frac{\frac{1}{2} \left( \frac{F_1}{m_1} - \mu g \right) t^2}{\frac{1}{2} \left( \frac{F_2}{m_2} - \mu g \right) t^2}
Simplifying:
S1S2=F1m1−μgF2m2−μg\frac{S_1}{S_2} = \frac{\frac{F_1}{m_1} - \mu g}{\frac{F_2}{m_2} - \mu g}
Since F1=F2F_1 = F_2, the ratio becomes:
S1S2=1m11m2=m2m1\frac{S_1}{S_2} = \frac{\frac{1}{m_1}}{\frac{1}{m_2}} = \frac{m_2}{m_1}
Final Answer:
The ratio of the displacements of the two toy-carts is:
S1S2=m2m1\frac{S_1}{S_2} = \frac{m_2}{m_1}
Thus, the correct answer is:
(A) S1S2=m2m1\frac{S_1}{S_2} = \frac{m_2}{m_1}.
