Question

A hockey ball at rest is hit by a stick, such that the force acts on the ball for 0.15s . If the ball is of mass 100 g and covers a distance of 100 m in 2 seconds, find the magnitude of the force applied by the hockey stick. Assume no friction is acting on the ball while rolling on the ground.


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Answer 1

→ Hey Mate,

→ Given Question:-

→  A hockey ball at rest is hit by a stick, such that the force acts on the ball for 0.15s . If the ball is of mass 100 g and covers a distance of 100 m in 2 seconds, find the magnitude of the force applied by the hockey stick. Assume no friction is acting on the ball while rolling on the ground.

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→ To Find:-

→  The magnitude of the force applied by the hockey stick?

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→ Given :-

→ Distance Covered ⇒ 100m

→ Time Taken            ⇒  2s

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→  Law Used:-

→ ∵ $\frac{Distance covered}{Time Taken}$                                                                                                                        

             

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→ Solution:-

⇒  Disctance covered by the Ball

⇒ = 100m

⇒ Time Taken                            

⇒     =  2s

⇒ $\frac{100m}{2s}$  = 50 ms$^-1$

⇒ Now ,

⇒ Initial Velocity ( u ) = 0

⇒ Final  Velocity  ( v ) = 50ms$^-1$

⇒ Time                   ( t ) = 0.15s

⇒ Acceleration producted on ball ⇒ (α) = ?

⇒ v = u + αt

⇒ 50 = 0 + α × 0.15

⇒ α = $\frac{50}{0.15}$ = 333.3ms$^-2$

⇒ Force acting on the ball ,

⇒ F = $\frac{100}{1000} (kg)$ × 333.3ms$^-2$

⇒ = 33.33N

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→ More information:-

→ This question is about hockey ball and stick.

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Answer 2

→ Given Question:-

→  A hockey ball at rest is hit by a stick, such that the force acts on the ball for 0.15s . If the ball is of mass 100 g and covers a distance of 100 m in 2 seconds, find the magnitude of the force applied by the hockey stick. Assume no friction is acting on the ball while rolling on the ground.

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$

→ To Find:-

→  The magnitude of the force applied by the hockey stick?

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$

→ Given :-

→ Distance Covered ⇒ 100m

→ Time Taken            ⇒  2s

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$

→  Law Used:-

→ ∵ $\red{\frac{Distance covered}{Time Taken}}$                                                                                                                        

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$

→ Solution:-

⇒  Disctance covered by the Ball

⇒ = 100m

⇒ Time Taken                            

⇒     =  2s

⇒ $\frac{100m}{2s}$  = 50 ms$^-1$

⇒ Now ,

⇒ Initial Velocity ( u ) = 0

⇒ Final  Velocity  ( v ) = 50ms$^-1$

⇒ Time                   ( t ) = 0.15s

⇒ Acceleration producted on ball ⇒ (α) = ?

⇒ v = u + αt

⇒ 50 = 0 + α × 0.15

⇒ α = $\frac{50}{0.15}$ = 333.3ms$^-2$

⇒ Force acting on the ball ,

⇒ F = $\frac{100}{1000} (kg)$ × 333.3ms$^-2$

⇒ = 33.33N

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$