Chapter 9: Q92P (page 254)

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?

Expert verified

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

Step by step solution

Mass of block 1, $m_{1} = 1.0 kg$ .
Mass of block 2, $m_{2} = 3.0 kg$ .
The initial velocity of block 1, $v_{1 j} = 1.7 m / s$ .
Speed of the center of mass is, role="math" localid="1661248917282" $v_{c o m} = 0 m / s$ .

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

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

$m_{1} v_{1 i} + m_{2} v_{2 i} = (m_{1} + m_{2}) v_{c o m}$

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

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

$(1.0 kg) (1.7 \frac{m}{s}) + (3.0 k g) v_{2 i} = [(1.0 kg) + (3.0 kg)] (0 \frac{m}{s}) (1.0 kg) (1.7 \frac{m}{s}) + (3.0 k g) v_{2 i} = 0$

Solving further as,

$(3.0 kg) v_{2 i} = - 1.7 kg \cdot \frac{m}{s} = \frac{- 1.7 kg \cdot \frac{m}{s}}{3.0 kg} |v_{2 i}| = 0.567 \frac{m}{s}$

Therefore, the initial speed of the other block is 0.57 m/s . Its direction is opposite to that of $v_{1}$ .

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