Shortcut Methods

Shortcut Methods and Tricks

  1. A 20 kg block is at rest on a horizontal surface. A force of 100 N is applied to the block for 5 seconds. Calculate the final velocity of the block.

Shortcut Method:

  • Use the equation of motion:

$$v = u + at$$

  • where
  • $$v$$ is the final velocity
  • $$u$$ is the initial velocity (which is 0 in this case)
  • $$a$$ is the acceleration
  • $$t$$ is the time
  • Substituting the values, we get:

$$v = 0 + (100/20) * 5$$

$$v = 25 m/s$$

Trick:

  • Here’s a quick trick to calculate the final velocity:
  • Simply multiply the force applied by the time taken and divide the result by the mass of the object.

  • In this case, 100 N × 5 s = 500 Ns, and 500 Ns / 20 kg = 25 m/s.

  1. A car of mass 1000 kg is moving with a velocity of 36 km/h. The driver applies the brakes and the car comes to rest in 10 seconds. Calculate the force applied by the brakes.

Shortcut Method:

  • Use the equation of motion:

$$v = u + at$$

  • Rearranging the equation, we get:

$$a = (v - u)/t$$

  • Substituting the values, we get:

$$a = (0 - 36 * 5/18) / 10$$

$$a = -10 m/s^2$$

  • Now, use Newton’s second law:

$$F = ma$$

  • Substituting the values, we get:

$$F = 1000 * (-10)$$

$$F = -10000 N$$

Trick:

  • To find the force applied by the brakes quickly, you can multiply the mass of the car by the change in velocity and then divide by the time taken.

  • In this case, the change in velocity is (0 – 36 km/h) = -36 km/h, which is equivalent to -10 m/s. So, the force applied by the brakes is 1000 kg × (-10 m/s) / 10 s = -10000 N.

  1. A ball of mass 2 kg is thrown vertically upwards with an initial velocity of 20 m/s. Calculate the maximum height reached by the ball.

Shortcut Method:

  • Use the equation of motion for vertical motion:

$$v^2 = u^2 + 2as$$

  • At the maximum height, the final velocity of the ball will be 0 m/s.

  • Substituting the values, we get:

$$0^2 = (20)^2 + 2(-9.8) * s$$

$$s = 20.41 m$$

Trick:

  • There’s a quick way to calculate the maximum height without using the equation.

  • Simply square the initial velocity, divide it by 2 times the acceleration due to gravity (9.8 m/s²), and that will give you the maximum height reached.

  • In this case, (20 m/s)² / (2 x 9.8 m/s²) = 20.41 m.

  1. A body of mass 10 kg is at rest on a horizontal surface. A force of 5 N is applied to the body for 2 seconds. Calculate the final velocity of the body.

Shortcut Method:

  • Use the equation of motion:

$$v = u + at$$

  • where
  • $$v$$ is the final velocity
  • $$u$$ is the initial velocity (which is 0 in this case)
  • $$a$$ is the acceleration
  • $$t$$ is the time
  • Substituting the values, we get:

$$v = 0 + (5/10) * 2$$

$$v = 1 m/s$$

Trick:

  • Here’s a quick trick to find the final velocity:

  • Multiply the force applied by the time taken and divide the result by the mass.

  • In this case, 5 N × 2 s = 10 Ns, and 10 Ns / 10 kg = 1 m/s.

  1. A cyclist of mass 50 kg is moving with a velocity of 10 m/s. The cyclist applies the brakes and the cycle comes to rest in 5 seconds. Calculate the force applied by the brakes.

Shortcut Method:

  • Use the equation of motion:

$$v = u + at$$

  • Rearranging the equation, we get:

$$a = (v - u)/t$$

  • Substituting the values, we get:

$$a = (0 - 10) / 5$$

$$a = -2 m/s^2$$

  • Now, use Newton’s second law:

$$F = ma$$

  • Substituting the values, we get:

$$F = 50 * (-2)$$

$$F = -1000 N$$

Trick:

  • To find the force applied by the brakes quickly, you can multiply the mass of the cyclist by the change in velocity and then divide by the time taken.

  • In this case, the change in velocity is (0 – 10 m/s) = -10 m/s. So, the force applied by the brakes is 50 kg × (-10 m/s) / 5 s = -1000 N.

  1. A ball of mass 1 kg is thrown vertically upwards with an initial velocity of 10 m/s. Calculate the maximum height reached by the ball.

Shortcut Method:

  • Use the equation of motion for vertical motion:

$$v^2 = u^2 + 2as$$

  • At the maximum height, the final velocity of the ball will be 0 m/s.

  • Substituting the values, we get:

$$0^2 = (10)^2 + 2(-9.8) * s$$

$$s = 5.1 m$$

Trick:

  • There’s a quick way to calculate the maximum height without using the equation.

  • Simply square the initial velocity, divide it by 2 times the acceleration due to gravity (9.8 m/s²), and that will give you the maximum height reached.

  • In this case, (10 m/s)² / (2 x 9.8 m/s²) = 5.1 m.



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