Question

Two blocks of masses1.0 kgand3.0 kgare connected by a spring and rest on a frictionless surface. They are given velocities toward each other such that the 1.0 kgblock travels initially at 1.7 m/stoward the center of mass, which remains at rest. What is the initial speed of the other block?

Short answer

The initial speed of the other block is 0.57 m/s .

Step-by-step solution

Understanding the given information

1. Mass of block 1,$m_1=1.0\mathrm{kg}$.
2. Mass of block 2,$m_2=3.0\mathrm{kg}$.
3. The initial velocity of block 1,$v_{1j}=1.7\mathrm{m}/\mathrm{s}$.
4. Speed of the center of mass is, role="math" localid="1661248917282" $v_{com}=0\mathrm{m}/\mathrm{s}$.

Concept and formula used in the given question

Using the formula of conservation of the momentum, we can find the initial speed of the other block.

Calculation for the initial speed of the other block

The collision, in this case, is inelastic, so from the conservation of momentum theorem, we can write,

$m_1{v}_{1i}+m_2{v}_{2i}=\left(m_1+m_2\right)v_{com}$

Here $v_{2i}$is the initial velocity of block 2.

Substitute the values in the above expression, and we get,

$$\begin{gathered} \left(1.0\mathrm{kg}\right)\left(1.7\frac{\mathrm{m}}{\mathrm{s}}\right)+\left(3.0kg\right)v_{2i}=\left[\left(1.0\mathrm{kg}\right)+\left(3.0\mathrm{kg}\right)\right]\left(0\frac{\mathrm{m}}{\mathrm{s}}\right) \\ \left(1.0\mathrm{kg}\right)\left(1.7\frac{\mathrm{m}}{\mathrm{s}}\right)+\left(3.0kg\right)v_{2i}=0 \end{gathered}$$

Solving further as,

$$\begin{gathered} \left(3.0\mathrm{kg}\right){\mathrm{v}}_{2\mathrm{i}}=-1.7\mathrm{kg}·\frac{\mathrm{m}}{\mathrm{s}} \\ =\frac{-1.7\mathrm{kg}·{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}}{3.0\mathrm{kg}} \\ \left|{\mathrm{v}}_{2\mathrm{i}}\right|=0.567\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Therefore, the initial speed of the other block is 0.57 m/s . Its direction is opposite to that of $v_1$.