Question

# A bullet of mass m1=20gm fired horizontally with a velocity of v=200m/s hits a wooden block of mass m2=500 gm (take g=9.8m/s2) resting on a horizontal plane as shown in the figure and the bullet remains embedded in the block after the impact. If the coefficient of friction between the surface in contact remains constant at 0.3, the distance the block will move before coming to rest is

A
5.03 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
10.06 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
20.12 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
100 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 10.06 mInitial momentum of bullet and block = final momentum of bullet and block =201000×200+0=(500+20)1000×V ∴ v=7.69 m/sec2 Now apply work energy principle (final kinetic energy) - (initial kinetic energy) = network done 12mv2−0=μ×mgx 12×(5201000)×7.622=0.3×5201000×9.81×x ∴ x=10.046 m≈10.05m

Suggest Corrections
0