WBJEE · Physics · Laws of Motion
A golf ball of mass \(50 \mathrm{gm}\) placed on a tee, is struck by a golf-club. The speed of the golf ball as it leaves the tee is \(100 \mathrm{~m} / \mathrm{s}\), the time of contact on the ball is \(0.02 \mathrm{~s}\). If the force decreases to zero linearly with time, then the force at the beginning of the contact is
- A \(100 \mathrm{~N}\)
- B \(200 \mathrm{~N}\)
- C \(250 \mathrm{~N}\)
- D \(500 \mathrm{~N}\)
Answer & Solution
Correct Answer
(D) \(500 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
Impulse \(=\) Area under F-t graph \(=\) change in linear momentum \[ =\frac{1}{2} \times F \times 0.02=\frac{50}{1000}(100-0) \Rightarrow F=500 \mathrm{~N} \]
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