Question

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 1.47 kg) and embeds itself in block 2 (mass 2.02 kg). The blocks end up with speeds v1 = 0.590 m/s and v2 = 1.45 ?

Answer 1

The velocity of the bullet as it enters block 1 is `1086m.s^(-1)` and the velocity of the bullet as it leaves block 1 is `838 m.s^(-1)`.

Explanation:

Diagram supplied with question:

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Solution
This solution will use the principle of conservation of momentum. First it will be applied to the very start and very end of the problem. Then it will be applied to just before the bullet enters block 1 and just after it leaves block 1.

To begin with I will define the terms I will use for the solution (sorry but this will be a long list):
`u_B` the initial velocity of the bullet (unknown).
`u_1` the initial velocity of block 1 (`0 m .s^(-1)`).
`u_2` the initial velocity of block 2 (`0 m .s^(-1)`).
`v_1` the final velocity of block 1 (`0.59 m .s^(-1)`).
`v_2` the final velocity of block 2 (`1.45 m .s^(-1)`).
`v_(B1)` the velocity of bullet after exiting block 1 (unknown).
`m_B` the mass of the bullet (`3.5g = 0.0025 kg`).
`m_1` the mass of block 1 (`1.47kg`).
`m_2` the mass of block 2 (`2.02kg`).

Total Initial Momentum
This is the total momentum before the bullet hits block 1.
`p_i = m_B×u_B`
Total Final Momentum
This is the total momentum after the bullet has struck block 2.
`p_f = m_1×v_1 + (m_2+m_B)×v_2`

From the conservation of momentum `p_i = p_f`.
`⇒ m_B×u_B = m_1×v_1 + (m_2+m_B)×v_2`

Rearrange this for the initial velocity of the bullet:
`⇒ u_B = (m_1×v_1 + (m_2+m_B)×v_2)/m_B`
`=(1.47×0.59 + (2.02+0.0035)×1.45)/0.0035=1086… m.s^(-1)`
`u_B=1086m.s^(-1)`

The velocity of the bullet as it enters block 1 is `1086m.s^(-1)`.

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Next write an expression for the total momentum after the bullet has left block 1 but before it enters block 2:
`p_1=m_B×v_(B1)+m_1×v_1`

The principle of conservation of momentum requires that `p_i = p_1` also.
`⇒ m_B×u_B=m_B×v_(B1)+m_1×v_1`

Rearrange for the velocity of the bullet after it leaves block 1 (`v_(B1)`):
`⇒ v_(B1)=(m_B×u_B - m_1×v₁)/m_B`
`=(0.0035 × 1086.… - 1.47 × 0.59)/0.0035 = 838.3… m.s^(-1)`
`v_(B1)=838 m.s^(-1)`

The velocity of the bullet as it leaves block 1 is `838 m.s^(-1)`.