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Gnoma [55]
3 years ago
9

In a velocity versus time graph, what does the slope of a line at any point indicate?

Physics
1 answer:
VladimirAG [237]3 years ago
5 0
The acceleration and speed
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A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters pe
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The final speed of the block after the collision with the obstacle is \boxed{3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}.

Further Explanation:

Given:

The mass of the block is 6.0\,{\text{kg}}.

The initial momentum of the block is 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}.

The impulse imparted by the obstacle is 10\,{\text{N}} \cdot {\text{s}}.

Concept:

The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.

According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.

The expression for the impulse momentum theorem is.

{p_f} - p{ & _i} = I               …… (1)                                    

Substitute 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} for {p_i} and - 10\,{\text{N}} \cdot {\text{s}} for I  in equation (1).

 \begin{aligned}{p_f} &= - 10\,{\text{N}} \cdot {\text{s}} + 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 20\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}

The final momentum of the block can be expressed as:

{p_f} = m{v_f}                   …… (2)                                  

Substitute 20\text{kg}\;\text{m/s} for {p_f} and 6.0\,{\text{kg}} for m in equation (2).

 \begin{aligned}20 &= 6 \times {v_f} \\ {v_f}&= \frac{{20}}{6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}

Thus, the final speed of the block after the collision with the obstacle is \boxed{3.33\;\text{m/s}}.

Learn More:

  1. Choose the 200 kg refrigerator. Set the applied force to 400 n (to the right) brainly.com/question/4033012
  2. With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward brainly.com/question/9719731
  3. Which of the following is an example of a nonpoint source of freshwater pollution brainly.com/question/1482712

Answer Details:

Grade: High School

Chapter: Impulse-momentum theorem

Subject: Physics

Keywords:  Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.

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