Question #117859

A bullet with speed 511 m/s and mass m = 5.00 g is shot into a stationary block of wood hanging a distance

h1 = 1.00 m above the ground, as shown in Figure 1. The block is hanging by a string of length r = 0.30 m. The

bullet is lodged in the block, and the block-bullet system swings so that the string makes an angle of θ = 60.0°

when the string breaks, as shown in Figure 2. The block-bullet system then falls to the ground a horizontal

distance d2 = 2.00 m from its position when the string breaks. Ignoring the effects of air resistance, determine

the mass, M of the block of wood.

h1 = 1.00 m above the ground, as shown in Figure 1. The block is hanging by a string of length r = 0.30 m. The

bullet is lodged in the block, and the block-bullet system swings so that the string makes an angle of θ = 60.0°

when the string breaks, as shown in Figure 2. The block-bullet system then falls to the ground a horizontal

distance d2 = 2.00 m from its position when the string breaks. Ignoring the effects of air resistance, determine

the mass, M of the block of wood.

Expert's answer

**Explanations & Calculations**

- Refer to the figure & consider the mass of the block to be
**M**kg. - Applying the conservation of linear momentum to find V (velocity of the system after the collision)

"\\qquad\\qquad\n\\begin{aligned}\n\\small 0.005kg\\times \\overrightarrow{511} ms^{-1} +0 \n &=\\small (0.005 + M) V\\\\\n\\small V &= \\small \\frac{2.555}{(0.005 +M)}\n\\end{aligned}"

- From down here to up where the line brakes, apply conservation of mechanical energy to find the velocity effective for the projectile.

(Here the height where the line breaks = **1+h** = "1+(\\small 0.3 -0.3\\cos60) = \\small \\bold{1.15m}" )

"\\qquad\\qquad\n\\begin{aligned}\n\\small E_k +E_p &= \\small E_{k1} +E_{p1}\\\\\n\\scriptsize \\frac{1}{2}(0.005+M)\\times\\frac{2.555^2}{(0.005+M)^2} \n+0 &= \\scriptsize\\frac{1}{2}(0.005+M)V_1^2 + (0.005+M)gh\\\\ \n\\small \\frac{2.555^2}{2(0.005+M)} &= \\small \\frac{1}{2}(0.005+M)(V^2+2gh)\\\\\n\\small (0.005+M)^2 &= \\small \\frac{2.555^2}{(V^2+2gh)} \\cdots(1) \n\\end{aligned}"

- Now to find V, apply s = ut + 0.5at
^{2}both horizontally & vertically down.

"\\qquad\\qquad\n\\begin{aligned}\n\\small s &= ut\\\\\n2 &= \\small V\\cos60\\times t\\\\\n\\small t&= \\small\\frac{4}{V}\\cdots(2) \n\\end{aligned}" "\\qquad\\qquad\n\\begin{aligned}\n\\small \\downarrow s &= ut+0.5gt^2\\\\\n \\small 1.15 &= \\small -V\\sin60\\times t+0.5\\times9.8\\times t^2\\\\\n \\small 1.15 &= \\small \\frac{-\\sqrt{3}Vt}{2}+0.5\\times9.8\\times t^2\\\\\n\n\n\\end{aligned}"

- Now solving for V yields,

"\\qquad\\qquad\n\\small V=\\begin{cases}\n4.122 ms^{-1}\\\\\n-4.122 ms^{-1} &\\text{negligible; negative}\n\\end{cases}"

- Now apply this value in (1) to calculate M.

"\\qquad\\qquad\n\\begin{aligned}\n\\small (0.005+M)^2 &= \\small \\frac{2.555^2}{(V^2+2gh)}\\\\\n\\small (0.005+M)^2 &= \\small \\frac{2.555^2}{\\bold{4.122}^2+2\\times9.8 \\times0.15}\\\\\n\n\n\\end{aligned}"

- Solving for M yields,

"\\qquad\\qquad\n\\small M=\\begin{cases}\n0.567kg\\\\\n-0.577kg &\\text{negligible; negative}\n\\end{cases}"

- Therefore, the mass of the block is
**M = 0.567 kg**

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