Question

A 1.8-kg block, initially at rest, slides down a frictionless ramp that is angled at 35º to the horizontal. At a point 0.45 m down the slope it collides with and sticks to a stationary block of mass 1.1 kg. The blocks then continue another 0.88 m down the ramp. How long does the whole event take? (For those of you wondering how a block is stationary on a frictionless ramp, it was projected up the ramp from below so it had no speed at the time of impact.)

Answer 1

The whole event takes 0.69 seconds.

Step-by-step explanation:

To calculate the time taken for the whole event, we need to calculate the time taken for the two blocks to slide down the ramp separately and then add those times together.

For the first block, we can use the equation for acceleration down an inclined plane:

a = g * sin(theta)

where g is the acceleration due to gravity and theta is the angle of the ramp. Plugging in the values, we get:

a = 9.8 * sin(35°) = 5.67 m/s²

Using the equation for distance travelled,

d = ut + (1/2) * at²

where d is the distance, u is the initial velocity (which is 0 since the block is at rest), a is the acceleration, and t is the time, we can rearrange the equation to solve for t:

t = sqrt(2d/a)

Plugging in the values,

t = sqrt(2 * 0.45 / 5.67) = 0.30 s

So, the first block takes 0.30 seconds to slide down the ramp.

For the second block, we can use the same equations, but with different values for the distance and acceleration.

d = 0.88 m

a = g * sin(theta) = 9.8 * sin(35°) = 5.67 m/s²

Plugging in the values, we get:

t = sqrt(2d/a) = sqrt(2 * 0.88 / 5.67) = 0.39 s

So, the second block takes 0.39 seconds to slide down the ramp.

Adding the times together,

total time = 0.30 + 0.39 = 0.69 s

Therefore, the whole event takes 0.69 seconds.