Question

4. A 200 g mass is attached to a spring of spring constant k. The spring is
compressed 15 cm from its equilibrium value. When released the mass reaches a
speed of 5 m/s. What is the spring constant (in N/m)?

5. A 34-g bullet traveling at 120m/s embeds itself in a wooden block on a smooth
surface. The block then slides toward a spring and collides with it. The block
compresses the spring (k=100 N/m) a maximum of 1.25 cm. Calculate the mass
of the block of wood.

6. If a force of 300N is exerted upon a 60 kg mass for 3 seconds, how much
impulse does the mass experience?

Answer 1

Answer on Question #64086-Physics-Mechanics-Relativity

4. A 200 g mass is attached to a spring of spring constant k. The spring is compressed 15 cm from its equilibrium value. When released the mass reaches a speed of 5 m/s. What is the spring constant (in N/m)?

Solution

From the conservation of energy:

$$\frac{m v^{2}}{2} = \frac{k x^{2}}{2}$$ $$k = \frac{m v^{2}}{x^{2}} = 0.2 \left(\frac{5}{0.15}\right)^{2} = 222 \frac{N}{m}.$$

5. A 34-g bullet traveling at 120 m/s embeds itself in a wooden block on a smooth surface. The block then slides toward a spring and collides with it. The block compresses the spring (k=100 N/m) a maximum of 1.25 cm. Calculate the mass of the block of wood.

Solution

The law of conservation of momentum:

$$m v = (m + M) V$$ $$V = v \frac{m}{m + M}$$

From the conservation of energy:

$$\frac{(m + M) V^{2}}{2} = \frac{k x^{2}}{2}$$ $$\frac{m^{2} v^{2}}{2 (m + M)} = \frac{k x^{2}}{2}$$ $$M = \frac{m^{2} v^{2}}{k x^{2}} - m = 0.034 \left(\frac{0.034}{100} \left(\frac{120}{0.0125}\right)^{2} - 1\right) = 1065 \, \text{kg}.$$

6. If a force of 300 N is exerted upon a 60 kg mass for 3 seconds, how much impulse does the mass experience?

Solution

The impulse is

$$I = F t = 300 \cdot 3 = 900 \, \text{Ns}.$$

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