Chapter 11: Q. 54 (page 290)

Two $500 g$ blocks of wood are $2.0 m$ apart on a frictionless table. A $10 g$ bullet is fired at $400 m / s$ toward the blocks. It passes all the way through the first block, then embeds itself in the second block. The speed of the first block immediately afterward is $6.0 m / s$ . What is the speed of the second block after the bullet stops in it?

Short Answer

Expert verified

The speed of the second block after the bullet stops in it is $1.96 m / s$

Step by step solution

Step 1. Given information

Mass of block, $m = 500 g$

Mass of bullet, $m_{b u l l e t} = 10 g$

The velocity of bullet, $v_{b u l l e t} = 400 m / s$

Step 2. For the first collision

Apply conservation of principle for collision,

$m_{b u l l e t} \times v_{b u l l e t, i n t i a l} + m_{b l o c k} \times v_{f i r s t b l o c k, i n i t i a l} = m_{b u l l e t} \times v_{b u l l e t, f i n a l} + m_{b l o c k} \times v_{f i r s t b l o c k, f i n a l} 0.01 k g \times 400 m / s + 0.5 k g \times 0 m / s = 0.01 k g \times v_{b u l l e t, f i n a l} + 0.5 k g \times 6 m / s v_{b u l l e t, f i n a l} = 100 m / s$

Step 3. For second collision

Apply conservation of principle for collision,

$m_{b u l l e t} \times v_{b u l l e t, f} + m_{b l o c k} \times v_{s e c o n d b l o c k, i n i t i a l} = (m_{b u l l e t} + m_{b l o c k}) v_{s e c o n d b l o c k, f i n a l} 0.01 k g \times 100 m / s + 0.5 k g \times 0 m / s = (0.01 + 0.5) v_{s e c o n d b l o c k, f i n a l} v_{s e c o n d b l o c k, f i n a l} = 1.96 m / s$

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Short Answer

Step by step solution

Step 1. Given information

Step 2. For the first collision

Step 3. For second collision

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