Question

An object of mass 1 kg travelling in a straight line with a velocity of 10 m s^–1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Answer 1

Mass of the object, m₁ = 1 kg 
The velocity of the object before the collision, v₁ = 10 m/s 
Mass of the stationary wooden block, m₂ = 5 kg 
The velocity of the wooden block before the collision, v₂ = 0 m/s 
∴ Total momentum before collision = m₁ v₁ + m₂ v₂ = 1 (10) + 5 (0) = 10 kg m s^-1 
It is given that after the collision, the object and the wooden block stick together. 
The total mass of the combined system = m₁ + m₂ 
The velocity of the combined object = v 
According to the law of conservation of momentum: 
Total momentum before collision = Total momentum after collision 
m₁v₁+ m₂v₂
= (m₁+ m₂) v₁ 
=(10) + 5 (0) 
= (1 + 5)v 
= v = \(\frac{10}{6}\)
= \(\frac{5}{3}\) m/s
The total momentum after collision = 10 and the velocity of a combined object is \(\frac{5}{3}\) m/s.

Answer 2

Before the Impact:
The object's mass: \(1 \, \text{kg}\)
The object's velocity: \(10 \, \text{m/s}\)
The wooden block's mass: \(5 \, \text{kg}\)
The wooden block is initially stationary, so its velocity is \(0 \, \text{m/s}\).

The total momentum just before the impact is the sum of the momenta of the two objects:
\(p_{\text{before}} = p_{\text{object}} + p_{\text{block}}\)
\(p_{\text{before}} = (1 \, \text{kg} \times 10 \, \text{m/s}) + (5 \, \text{kg} \times 0 \, \text{m/s})\)
\(p_{\text{before}} = 10 \, \text{kg m/s} + 0 \, \text{kg m/s}\)
\(p_{\text{before}} = 10 \, \text{kg m/s}\)

After the Impact:
The two objects stick together and move off with a common velocity v'.
The total mass of the combined object: \(1 \, \text{kg} + 5 \, \text{kg} = 6 \, \text{kg}\).
Using the principle of conservation of momentum, the total momentum just after the impact is the same as before the impact:
\(p_{\text{before}} = p_{\text{after}}\)

So, the velocity of the combined object after the collision v' can be found using:
\(p_{\text{before}} = p_{\text{after}}\)
\(10 \, \text{kg m/s} = 6 \, \text{kg} \times v'\)
\(v' = \frac{10 \, \text{kg m/s}}{6 \, \text{kg}}\)

\(v' = \frac{5}{3} \, \text{m/s}\)

So, the answer is total momentum just before and after the impact: \(10 \, \text{kg m/s}\) and velocity of the combined object after the collision: \(\frac{5}{3} \, \text{m/s}\)