Math Problem Statement

A small dart of mass 0.020 kg is launched at an angle of 30° with an initial speed of 10 m/s. It collides with a block of mass 0.10 kg suspended by a string 1.2 m long. After the dart strikes, it sticks to the block, and they swing together.

Solution

Let's break down the problem step by step based on the image you've uploaded.

Problem Overview:

A dart of mass 0.020 kg is launched at a 30° angle with an initial speed of 10 m/s. The dart collides with a block of mass 0.10 kg, which is suspended on a string (length 1.2 m). After collision, the dart sticks to the block, and the two swing together until they make an angle θ with the vertical.

Here are the parts:

(a) Determine the speed of the dart just before it strikes the block.

For this, we need to resolve the horizontal and vertical components of the dart's velocity just before impact. From the image, it looks like you're using energy or kinematic principles. Let's confirm and calculate.

  • Initial velocity of the dart: v0=10m/sv_0 = 10 \, \text{m/s}
  • Angle of projection: θ=30\theta = 30^\circ

The speed just before impact (horizontal component remains constant): vdart, before impact=v0cos(θ)=10cos(30)=1032=8.66m/sv_{\text{dart, before impact}} = v_0 \cos(\theta) = 10 \cdot \cos(30^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 8.66 \, \text{m/s}

(b) Calculate the horizontal distance dd between the launching point and the point directly below the block.

To calculate the range, you're using the formula: R=v02sin(2θ)gR = \frac{v_0^2 \sin(2\theta)}{g} Substituting values: R=102sin(60)9.8=100329.8=88.37mR = \frac{10^2 \cdot \sin(60^\circ)}{9.8} = \frac{100 \cdot \frac{\sqrt{3}}{2}}{9.8} = 88.37 \, \text{m}

Thus, the horizontal distance is approximately 88.37 m.

(c) Calculate the speed of the block just after the dart strikes.

The law of conservation of momentum applies here. Before the collision, only the dart has momentum. After the collision, both the dart and block move together with a common velocity vfv_f.

Using conservation of momentum: m1v1+m2v2=(m1+m2)vfm_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f Here, v2=0v_2 = 0 (the block is initially at rest), and we solve for vfv_f: 0.0210=(0.02+0.10)vf0.02 \cdot 10 = (0.02 + 0.10) v_f 0.2=0.12vf0.2 = 0.12 v_f vf=0.20.12=1.67m/sv_f = \frac{0.2}{0.12} = 1.67 \, \text{m/s}

Thus, the speed of the block and dart after collision is approximately 1.67 m/s.


Would you like more details or clarifications on any part of the solution?

Here are 5 related questions you can explore next:

  1. How high will the block rise after the collision?
  2. What angle does the block-dart system make with the vertical at maximum height?
  3. What is the tension in the string just after the dart strikes the block?
  4. How much mechanical energy is lost in the collision?
  5. What would change if the block were moving horizontally before impact?

Tip: Always check for energy conservation after inelastic collisions to account for potential energy losses.

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Conservation of Momentum
Trigonometry
Physics Kinematics

Formulas

v = v_0 * cos(θ)
R = (v_0^2 * sin(2θ)) / g
m1 * v1 + m2 * v2 = (m1 + m2) * v_f

Theorems

Conservation of Linear Momentum
Projectile Motion Equations

Suitable Grade Level

Grades 10-12 (High School Physics)