Question

A 12.0-g bullet is fired horizontally into a 116-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 151 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 81.0 cm, what was the speed of the bullet at impact with the block?

Answer #1

**Solution:-**

**Given -**

Mass of bullet M1 = 12.0g = 0.012kg

Mass of block M2 = 116g = 0.116kg

Spring constant k = 151N/m

Compression in spring x = 81.0cm = 0.81m

Let, **V0 – Speed of bullet at impact**

V – Common speed of bullet + block after impact

Consider the, Bullet + the block + the spring as a system

From the time the bullet gets embedded in the block to the time when the spring undergoes maximum compression, the mechanical energy of the system is conserved.

Therefore,

**½*(M1+M2)V ^{2} = ½ Kx^{2}**

Or M1+M2)V^{2} = Kx^{2}

Or V^{2} = Kx^{2}/ M1+M2

Or V = x √K/ M1+M2

V = 0.81 √ 151/(0.012+0.116)

**V = 27.82 m/s**

Momentum of the system is conserved during the impact therefore

**M1V0 = (M1+M2)V**

V0 = (M1+M2)V /M1

V0 = (0.012+0.116)27.82 / 0.012

**V0 = 296.74 m/s**

This is the speed of the bullet at the impact with the block.

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