A rubber ball of mass $50 g m$ falls from a height of $1 m$ and rebounds to a height of $0.5 m$ . Find the impulse and the average force between the ball and the ground if the time for which they were in contact is $0.1 s .$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$3.8 \hat{j} N s, 3.8 N$

$3.8 \hat{j} N s, 0.38 N$

None of the above

$0.38 \hat{j} N s, 3.8 N$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$V e l o c i t y o f b a l l j u s t b e f o r e c o l l i s i o n; v^{2} = 0 + 2 g h v = v_{i} = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 1} M o m e n t u m o f b a l l b e f o r e c o l l i s i o n, \vec{P_{i}} = m v_{i} = 0.050 \times (- 4.43) \hat{j} N s = - 0.22 \hat{j} N s V e l o c i t y o f b a l l j u s t a f t e r c o l l i s i o n, V_{f} = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 0.5} = 3.13 m / s M o m e n t u m o f b a l l j u s t a f t e r c o l l i s i o n, \vec{P_{f}} = m v_{f} = 0.050 \times (3.13 \hat{j}) = 0.16 \hat{j} N s N o w i m p u l s e i m p a r t e d b y g r o u n d o n t h e b a l l \vec{J} = \vec{P_{f}} - \vec{P_{i}} = 0.16 \hat{j} - (- 0.22 \hat{j}) = 0.38 \hat{j} N s$