when a 0.05kg bullet is fired into a block that is suspended by a long cord so that it can swing as a pendulum. if the block is displaced so that its centre of gravity rises by 0.10m,what was the speed of the bullet?

Need to know mass of the block to do conservation of momentum.

565

(281.4m/s)

To solve this problem, we need to use the principle of conservation of mechanical energy. The mechanical energy of the system (bullet + block) is conserved if we neglect any dissipative forces such as air resistance and friction.

1. First, let's calculate the potential energy gained by the block when it is displaced:
Potential energy (PE) = mass (m) × gravitational acceleration (g) × height (h)
PE = 0.05 kg × 9.8 m/s^2 × 0.10 m = 0.049 J

2. Next, we need to calculate the kinetic energy of the bullet just before it hits the block. To do this, we'll need the mass of the bullet and its velocity.
However, this information is not provided in the question, so we need to make an assumption or find another piece of information to calculate it.

3. If we assume that the bullet was initially at rest (i.e., its initial velocity is 0 m/s), we can calculate its kinetic energy just before impact. In this case, the initial mechanical energy of the system is only due to the displacement of the block.
Therefore, kinetic energy (KE) of the bullet = potential energy of the block
KE = 0.049 J

4. The kinetic energy of the bullet is given by the formula:
KE = (1/2) × mass × velocity^2

5. Rearranging the formula, we can solve for velocity:
velocity = sqrt((2 × KE) / mass)

6. Plugging in the values:
velocity = sqrt((2 × 0.049 J) / 0.05 kg)
velocity ≈ 0.44 m/s

Therefore, the speed of the bullet just before it hits the block is approximately 0.44 m/s.