Math

Question\begin{tabular}{|c|c|} \hline 5 & \begin{tabular}{l} A bullet of mass 0.10 kg travels horizontally at a speed of 600 m/s600 \mathrm{~m} / \mathrm{s}. It strikes a stationary wooden block of mass 1.90 kg resting on a frictionless, horizontal surface. \\ The bullet stays in the block. \\ What is the speed of the bullet and the block immediately after the impact? \\ A 30 m/s30 \mathrm{~m} / \mathrm{s} \\ B 32 m/s32 \mathrm{~m} / \mathrm{s} \\ C 60 m/s60 \mathrm{~m} / \mathrm{s} \\ D 134 m/s134 \mathrm{~m} / \mathrm{s} \end{tabular} \\ \hline 6 & \begin{tabular}{l} A moving body undergoes a change of momentum. \\ What is a unit for change of momentum? \\ A Nm \\ B N/m\mathrm{N} / \mathrm{m} \\ C Ns \\ D N/s \end{tabular} \\ \hline 7 & \begin{tabular}{l} The diagram shows a uniform bridge, 4.0 m long and weighing 10000 N . \\ The bridge is pivoted at one end. A force at the other end gradually increases until the bridge begins to lift. \\ What is the lifting force as the bridge starts to move upwards? \\ A 2500 N \\ B 5000 N \\ C 10000 N \\ D 20000 N \end{tabular} \\ \hline 8 & \begin{tabular}{l} The diagram shows a wooden beam of weight 20 N . The centre of mass of the beam is labelled MM. \\ There is a pivot at one end of the beam. The beam is kept horizontal by an upward force, FF. \\ What is the magnitude of FF ? \\ A 12 N\quad 12 \mathrm{~N} \\ B 20 N \\ C 30 N \\ D 33 N \end{tabular} \\ \hline \end{tabular}

Studdy Solution

STEP 1

What is this asking? A bullet embeds itself in a block of wood, and we need to find their combined speed afterwards. Watch out! Momentum is key here, and remember that the bullet and block move together *after* the impact.

STEP 2

1. Momentum before the impact
2. Momentum after the impact
3. Find the final velocity

STEP 3

Alright, let's **calculate** the momentum of the bullet *before* it hits the block.
Remember, momentum is mass times velocity!
The bullet's mass is 0.100.10 kg, and its velocity is a blazing 600600 m/s.

STEP 4

So, the bullet's initial momentum is \(0.10 \text{ kg} \cdot 600 m/s600 \text{ m/s} = 60 kgm/s60 \text{ kg} \cdot \text{m/s}).
That's our **starting momentum**.
The block is stationary, so its momentum is 1.90 kg0 m/s=0 kgm/s1.90 \text{ kg} \cdot 0 \text{ m/s} = 0 \text{ kg} \cdot \text{m/s}.
The total momentum before impact is 60 kgm/s+0 kgm/s=60 kgm/s60 \text{ kg} \cdot \text{m/s} + 0 \text{ kg} \cdot \text{m/s} = 60 \text{ kg} \cdot \text{m/s}.

STEP 5

Now, the bullet and block are stuck together!
They're basically one big mass now.
This combined mass is 0.10 kg+1.90 kg=2.00 kg0.10 \text{ kg} + 1.90 \text{ kg} = 2.00 \text{ kg}.
Let's call their **final velocity** after the impact vv.

STEP 6

The momentum after impact is the combined mass multiplied by this **final velocity** vv: \(2.00 \text{ kg} \cdot v\).

STEP 7

Here's the magic of momentum conservation: the total momentum *before* the impact must equal the total momentum *after* the impact.
No momentum is lost!

STEP 8

So, we can set up a nice little equation: 60 kgm/s=2.00 kgv60 \text{ kg} \cdot \text{m/s} = 2.00 \text{ kg} \cdot v.

STEP 9

To find vv, we **divide** both sides of the equation by 2.00 kg2.00 \text{ kg}.
This gives us v=60 kgm/s2.00 kg=30 m/sv = \frac{60 \text{ kg} \cdot \text{m/s}}{2.00 \text{ kg}} = 30 \text{ m/s}.

STEP 10

The final speed of the block and bullet together is 3030 m/s.
That's answer A!

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