Question

A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters per second, strike an obstacle. The obstacle exerts an impulse of 10.0 newton · seconds to the west on the block. What is the speed of the block after the collision?

Answer 1

Answer:

33.3 m/s

Explanation:

Assuming eastward as positive direction and westward as negative direction, the initial momentum of the block is equal to:

$p_i = +30.0 kg m/s$

The surface is frictionless, so this momentum is conserved until the block hits the obstacle.

The obstacle exerts an impulse of

$I=-10.0 Ns$

toward west.

Due to the law of conservation of momentum, this impulse is equal to the variation of momentum of the block:

$\Delta p= p_f -p_i = I$

So we can find the final momentum of the block:

$p_f = p_i +I=30 kg m/s - 10.0 kg m/s=20 kg m/s$

Since we know the mass of the block, m = 6.0 kg, the final speed of the block will be

$v=\frac{p_f}{m}=\frac{20.0 kg m/s}{6.0 kg}=+3.33 m/s$

Answer 2

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:

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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.