A 11.0-g bullet is fired horizontally into a 101-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 160 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 76.0 cm, what was the speed of the bullet at impact with the block?
energy in spring=1/2 kx^2 compute that.
Intiial velocity of block/bullet combo:
= 1/2 (M+m)V=energy in spring, solve for V
Now, using the conservation of momentum
momentumbullet=momentum(block+bullet)
mVb=(m+M)V solve for Vb
To find the speed of the bullet at impact with the block, we can use the principles of conservation of momentum and conservation of mechanical energy. Here's how you can approach the problem:
1. Calculate the momentum of the bullet before the collision:
- The momentum (p) of an object is given by the product of its mass (m) and velocity (v): p = m * v.
- The mass of the bullet is 11.0 g, which is equivalent to 0.011 kg.
- Since the bullet is fired horizontally, its initial vertical velocity is 0 m/s.
- Thus, the momentum before the collision can be calculated as: p_bullet_before = m_bullet * v_bullet_horizontal.
2. Calculate the velocity of the bullet-block system after the collision:
- Since the bullet becomes embedded in the block, the mass of the block-bullet system (m_system) is the sum of the bullet mass and the block mass (101 g, which is equivalent to 0.101 kg).
- By the law of conservation of momentum, the momentum after the collision (p_system_after) is equal to the momentum before the collision (p_bullet_before): p_system_after = p_bullet_before.
- Calculate the velocity of the bullet-block system after the collision using the combined mass of the system and the calculated momentum: v_system_after = p_system_after / m_system.
3. Calculate the displacement of the block-bullet system:
- The displacement of the block-bullet system is given as 76.0 cm, which is equivalent to 0.76 m.
- The block-bullet system is connected to a spring, so we can use Hooke's Law to relate the displacement and the force exerted by the spring: F_spring = k * x, where k is the spring constant and x is the displacement.
- Rearrange the equation to find the force: F_spring = k * x.
- The force exerted by the spring is responsible for stopping the block-bullet system and compressing the spring, so we can equate the force exerted by the spring to the force of the block-bullet system: F_spring = m_system * a, where a is the acceleration of the block-bullet system.
- Rearrange the equation to find the acceleration: a = F_spring / m_system.
- Solve for the acceleration using the given values of the spring constant and the displacement: a = F_spring / m_system = (k * x) / m_system.
4. Relate the displacement and the velocity to calculate the spring constant:
- The potential energy stored in a spring (U_spring) is given by the equation: U_spring = (1/2) * k * x^2, where k is the spring constant and x is the displacement.
- The maximum potential energy stored in the compressed spring is equal to the maximum kinetic energy of the block-bullet system before it comes to a stop.
- At maximum compression, all the kinetic energy (KE_initial) of the block-bullet system is stored as potential energy: KE_initial = U_spring_max.
- Substitute the equations for kinetic and potential energy and solve for the spring constant: (1/2) * m_system * v_system_after^2 = (1/2) * k * x^2.
5. Substitute the values and solve for v_bullet_horizontal:
- Use the obtained spring constant (k) and displacement (x) in the equation from step 4 to calculate the spring constant.
- Substitute the values into the equation and solve for v_bullet_horizontal: m_bullet * v_bullet_horizontal = (1/2) * m_system * v_system_after^2.
By following these steps, you should be able to find the speed of the bullet at impact with the block.