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 and acceleration of 2.5×104m/s2. If the acceleration is constant, what impulse in 10−10kgm/s is delivered to a pollen of mass 1.0×10−7g? (do not include unit in answer)

Answer 1

Answer: $7.5(10)^{-10} kg m/s$

Explanation:

The impulse $p$ is given by:

$p=mV$ (1)

Where:

$m=1(10)^{-7}g \frac{1 kg}{1000 g}=1(10)^{-10} kg$ is the mass of the pollen

$V$ is the velocity of the pollen

On the other hand, we know acceleration $a$ is:

$a=\frac{V}{t}$ (2)

Where $t=0.3 ms=0.3(10)^{-3} s$ is the time and $a=2.5(10)^{4}m/s^{2}$

Then: $V=a.t$ (3)

Substituting (3) in (1):

$p=m.a.t$ (4)

$p=(1(10)^{-10} kg)(2.5(10)^{4}m/s^{2})(0.3(10)^{-3} s)$ (5)

Finally:

$p=7.5(10)^{-10} kg m/s$

Answer 2

Impulse : 7.5.10⁻¹⁰N.s

Further explanation

Momentum is the product of the mass of an object and its velocity

p = mv

Whereas impulse is defined is the product of the force with the time interval when the force is acting on the object.

I = F. Δt

the pollen grains to be ejected out of the flower in 0.30 ms and acceleration of 2.5 × 10⁴ m/s², then

Δt = 0.3 ms = 3.10⁻⁴ s

a = 2.5 × 10⁴ m/s²

mass of pollen: 1.0 × 10⁻⁷g = 1.0 x 10⁻¹⁰ kg

then:

A force that works according to Newton's second law:

F = m. a

$\rm F=1.0\times 10^{-10}\times 2.5\times 10^4\\\\F=2.5\times 10^{-6}\:N$

Impulse (I) :

$\rm I=2.5 \times 10^{-6}\times 3\times 10^{-4}\\\\I=7.5\times 10^{-10}N.s$

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