A bullet mass 10 g strikes a log with a mass of 25 g suspended from a 3 meter long string. The bullet lodges in the log which swings up through an angle of 30 degrees. Find the speed of the bullet just before it hit the log.
The law of conservation of linear momentum
m•v = (m +M) •u.
v =(m +M) •u/m.
The law of conservation of energy
KE =PE,
(m+M) •u^2/2 =(m+M) • g•h,
u = sqrt (2•g•h).
Since h =L(1-cosα) ,
v =[(m +M) /m] • sqrt[2•g•L(1-cosα)].
To find the speed of the bullet just before it hit the log, we can use the principle of conservation of momentum.
1. Start by finding the initial momentum of the bullet and log system. Momentum is defined as the product of mass and velocity. Given that the bullet has a mass of 10 g (0.01 kg) and the log has a mass of 25 g (0.025 kg), we can calculate the initial momentum as follows:
Initial momentum = mass of the bullet × velocity of the bullet + mass of the log × velocity of the log
As the log is initially at rest, its velocity is zero. Thus, the equation simplifies to:
Initial momentum = mass of the bullet × velocity of the bullet
2. Next, we need to calculate the final momentum of the bullet and log system. After the bullet lodges in the log, the system will swing up, and we are given that the log swings through an angle of 30 degrees. At this point, we can assume the system reaches its highest point, where the velocity of the bullet becomes zero.
Final momentum = mass of the bullet × 0 (velocity of the bullet at the highest point) + mass of the log × velocity of the log (velocity of the log at the highest point)
Since the log reaches its highest point and momentarily stops before reversing its direction, the velocity of the log at the highest point is also zero.
Final momentum = 0 (mass of the bullet × 0) + mass of the log × 0
3. According to the principle of conservation of momentum, the initial momentum and the final momentum of a system should be equal. Hence:
Initial momentum = Final momentum
mass of the bullet × velocity of the bullet = 0 + 0
4. Rearrange the equation to solve for the velocity of the bullet:
velocity of the bullet = 0 / mass of the bullet
velocity of the bullet = 0 m/s
Therefore, the speed of the bullet just before it hit the log is zero.