Question

The flowers of the bunchberry plant open with astonishing force and speed, causing the pollen grains to be ejected out of the flower in a mere 0.30 ms at an acceleration of 2.5 × 104 m. s2 If the acceleration is constant, what impulse is delivered to a pollen grain with a mass of 1.0 × 10−7g?

Answer 1

Answer:

I = 7.5*10^-10 kg m/s

Explanation:

In order to calculate the impulse you first take into account the following formula:

$I=m\Delta v=m(v-v_o)$       (1)

m: mass of the pollen grain = 1.0*10^-7g = 1.0*10^-10 kg

v: final speed of the pollen grain = ?

vo: initial speed of the pollen grain = 0 m/s

Next, you calculate the final speed of the pollen grain by using the information about the acceleration and time. You use the following formula:

$v=v_o+a t$       (2)

a: acceleration = 2.5*10^4 m/s^2

t: time = 0.30ms = 0.30*10^-3 s

$v=0m/s+(2.5*10^4m/s^2)(0.30*10^{-3}s)=7.5\frac{m}{s}$

Next, you replcae this value of v in the equation (1) and calculate the impulse:

$I=m(v-v_o)=(1.0*10^{-10}kg)(7.5m/s-0m/s)=7.5*10^{-10}kg.\frac{m}{s}$

The impulse delivered to the pollen grain is 7.5*10^-10 kg m/s