The diagram shows a bullet moving towards a wooden block - Edexcel - GCSE Physics - Question 4 - 2013 - Paper 1

To calculate the velocity after the collision, we will apply the principle of conservation of momentum:

$$ext{momentum before} = ext{momentum after}$$

Calculating the total momentum before the collision:

$$ext{momentum before} = 5.1 \, ext{kg m/s} ext{ (from bullet)} + 0 \, ext{kg m/s} ext{ (from block)} = 5.1 \, ext{kg m/s}$$

After the collision, the combined mass is:

$$ext{mass after} = 0.030 \, ext{kg} + 0.80 \, ext{kg} = 0.83 \, ext{kg}$$

Now, applying the conservation of momentum:

$$5.1 \, ext{kg m/s} = 0.83 \, ext{kg} imes v$$

Solving for the velocity $v$:

v = rac{5.1 \, ext{kg m/s}}{0.83 \, ext{kg}} \ = 6.14 \, ext{m/s} ext{ (approximately 6.1 m/s)}

Thus, the velocity of the wooden block and bullet immediately after the collision is approximately 6.1 m/s.