6 (a) An atom of mass 6.6 × 10^{-27} kg is moving with a velocity of 480 m/s. Calculate the momentum of the atom. (b) Figure 11 shows a ball before and after it collides with a wall. The arrows show the direction of movement of the ball. Before the collision, the momentum of the ball is 0.80 kg m/s. After the collision, the momentum of the ball is 0.60 kg m/s in the opposite direction. The ball is in contact with the wall for a time of 70 ms during the collision. Calculate the force exerted on the ball by the wall. Use an equation selected from the list of equations at the end of the paper.

Photo AI

6 (a) An atom of mass 6.6 × 10^{-27} kg is moving with a velocity of 480 m/s - Edexcel - GCSE Physics: Combined Science - Question 6 - 2023 - Paper 1

Question 6

$6-(a)-An-atom-of-mass-6.6-×-10^{-27}-kg-is-moving-with-a-velocity-of-480-m/s-Edexcel-GCSE Physics: Combined Science-Question 6-2023-Paper 1.png$

6 (a) An atom of mass 6.6 × 10^{-27} kg is moving with a velocity of 480 m/s. Calculate the momentum of the atom. (b) Figure 11 shows a ball before and after it col... show full transcript

Worked Solution & Example Answer:6 (a) An atom of mass 6.6 × 10^{-27} kg is moving with a velocity of 480 m/s - Edexcel - GCSE Physics: Combined Science - Question 6 - 2023 - Paper 1

Step 1

Calculate the momentum of the atom.

96%

114 rated

Answer

To find the momentum (p) of the atom, we can use the formula:

$p = m imes v$

where:

$m$ is the mass of the atom, and
$v$ is the velocity of the atom.

Substituting the given values: $p = 6.6 imes 10^{-27} ext{ kg} imes 480 ext{ m/s}$

Calculating this gives: $p = 3.168 imes 10^{-24} ext{ kg m/s}$

Step 2

Calculate the force exerted on the ball by the wall.

99%

104 rated

Answer

To calculate the force (F) exerted by the wall on the ball, we can use the formula:

$F = \frac{\Delta p}{\Delta t}$

where:

$\Delta p$ is the change in momentum, and
$\Delta t$ is the time of contact.

First, we calculate the change in momentum: $\Delta p = p_{final} - p_{initial} = (0.60 ext{ kg m/s}) - (0.80 ext{ kg m/s}) = -0.20 ext{ kg m/s}$

Next, we use the time of contact, which is given as 70 ms (or $70 imes 10^{-3}$ seconds):

Now substituting in the values: $F = \frac{-0.20 ext{ kg m/s}}{70 \times 10^{-3} ext{ s}}$

Calculating this gives: $F = -2.857 ext{ N}$ (The negative sign indicates the force is in the opposite direction of the initial momentum of the ball.)

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered