Please make sure this is right!!Mechanic Physics:In Figure (1), a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block 1 (mass 1.05 kg) and embeds itself in block 2 (mass 1.76 kg). The blocks end up with speeds v1 = 0.510 m/s and v2 = 1.40 m/s (see Figure (2)). Neglecting the material removed from block 1 by the bullet, find the speed of the bullet as it (a) leaves and (b) enters block 1. The image is a visual portrayal of a physics problem focusing on the conservation of momentum in a frictionless environment.### Description:#### Diagram (1):- **Initial State:** - There are two blocks, labeled "1" and "2," positioned on a frictionless horizontal surface. - A small projectile is depicted moving towards block "1" from the left-hand side. - The surface is noted to be frictionless, allowing for a smooth motion of the blocks without resistance.#### Diagram (2):- **Final State:** - After the collision, the projectile has presumably struck block "1", causing both blocks to move. - Block "1" is now moving to the right with velocity denoted as "v1." - Block "2" is also moving to the right with velocity denoted as "v2." - The dashed arrows within each block indicate the direction and relative magnitude of their velocities.### Analysis:- This scenario can be analyzed using the principles of linear momentum conservation, which states that in the absence of external forces, the total momentum of a system remains constant.- Considering the frictionless surface, the only interactions that change the momentum of the blocks are the collision with the projectile.### Equations:- Let the mass of block "1" be \( m_1 \), the mass of block "2" be \( m_2 \), the mass of the projectile be \( m_p \), and the initial velocity of the projectile be \( u_p \). - The initial momentum of the system is given by \( m_p \times u_p \).- The final momentum of the system is the sum of the momenta of the two blocks and the projectile post-collision: \( m_1 \times v_1 + m_2 \times v_2 \).- By the law of conservation of momentum: \[ m_p \times u_p = m_1 \times v_1 + m_2 \times v_2 \]This setup provides the fundamental basis for further calculations and analysis in physics problems related to collisions and momentum conservation.

Name: Please make sure this is right!!Mechanic Physics:In Figure (1), a 3.50
Uploaded: 2024-03-20T06:57:37.000Z
Channel: Bartleby
Description: Please make sure this is right!!Mechanic Physics:In Figure (1), a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block 1 (mass 1.05 kg) and embeds itself in block 2 (mass 1.76 kg). The blo