Question

Two spheres of mass 10g and 100g are dropped to two pans from height 40 cm and 10 cm respectively.If,The two balls come to rest after 0.1 second,Calculate the force exerted by each sphere on the pans.

Answer 1

Answer

• Force exerted by 10 g sphere is 2.82 N
• Force exerted by 100 g sphere is 1.41 N

Given

• Two spheres of mass 10 g and 100 g are dropped to two pans from height 40 cm and 10 cm respectively.If,The two balls come to rest after 0.1 second

To Find

• The force exerted by each sphere on the pans

Formula Used

$\bigstar \;\;\; \bf v^2-u^2=2as\\\\\bigstar \;\;\; \bf F=ma$

Solution

For Body 1 :

Mass , m = 10 g = 0.01 kg

Initial velocity , u = 0 m/s [ Since dropped ]

Distance = Height , d = 40 cm = 0.4 m

Time , t = 0.1 s

Final velocity , v = ? m/s

Acceleration , a = ? m/s²

Apply 3 rd equation of motion .

⇒ v² - u² = 2as

⇒ v² - 0² = 2(10)(0.4)

⇒ v² =

⇒ v = 2.82 m/s

So ,

Final velocity , v = 2.82 m/s

Now ,

$\rm F=ma\\\\\implies \rm F=m\bigg(\dfrac{v-u}{t}\bigg)\\\\\implies \rm F=0.01\bigg(\dfrac{2.82-0}{0.1}\bigg)\\\\\implies \rm F=2.82\ N$

For Body 2 :

Mass , m = 100 g = 0.1 kg

Initial velocity , u = 0 m/s [ Since dropped ]

Distance = Height , d = 10 cm = 0.1 m

Time , t = 0.1 s

Acceleration , a = ? m/s²

Final velocity , v = ? m/s

Apply 3rd equation of motion .

⇒ v² - u² = 2as

⇒ v² - 0² = 2(10)(0.1)

⇒ v² = 20(0.1)

⇒ v² = 2

⇒ v = 1.41 m/s

So ,

Final velocity , v = 1.41 m/s

Now ,

$\rm F=ma\\\\\implies \rm F=m\bigg(\dfrac{v-u}{t}\bigg)\\\\\implies \rm F=0.1\bigg(\dfrac{1.41-0}{0.1}\bigg)\\\\\implies \rm F=1.41\ N$