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An object of mass $1 kg$ travelling in a straight line with a velocity of $10 m / s$ collides with, and sticks to, a stationary wooden block of mass $5 kg$ . Then, they both move off together in the same straight line. Velocity of the combined object is ____ m/s.

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Detailed Solution

Velocity of the combined object is

1.67 m / s .

Here, mass of object,

m₁ = 1 kg

Velocity of object,

v₁ = 10 m / s

Total momentum just before the impact

= m₁ v₁

\Rightarrow 1 kg \times 10 m / s

\Rightarrow 10 kg . m / s

From the principle of conservation of momentum
Total momentum just after the impact = Total momentum just before the impact

\Rightarrow

10 kg m/s
Now, this object of mass

1 kg

sticks to a stationary wooden block of mass

5 kg

. So,
Mass of object + mass of block

\Rightarrow 1 kg + 5 kg

\Rightarrow 6 kg

Velocity of combined object =?
According to laws of conservation of momentum.
m₁v1 = m2v2
v2 =

\frac{m₁v₁}{m 2}

\Rightarrow

\frac{1 kg \times 10 m / s}{6 kg}

\Rightarrow

1.67 m / s

Velocity of the combined object is

1.67 m / s

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