A build of mass ml = k.050 kg travelling at 200 m/s, strikes and passes through
ID: 2019992 • Letter: A
Question
A build of mass ml = k.050 kg travelling at 200 m/s, strikes and passes through a block of mass m2 = 4.0 kg emerging with a velocity of 50 m/s The block is suspended vertically by a 1 in long string . Is kinetic energy conserved in the impact? Is momentum conserved in the impact? What is the velocity of the block IMMEDIATELY after the impact? How high does the block use at the highest point? What is the angle, then, between the string and the vertical What would the velocity of the bullet have to be for the block to rise so the angle between the string and the vertical to be 90 degree (assume that the bullet still passes through the block and leaves with a Velocity of 50 m/s)Explanation / Answer
Given that the mass of bullet is m1 = 0.050 kg Initial velocity of the bullet is u1 = 200m/s mass of block is m2 = 4.0 kg Final velocity of the bullet after collision is u2 = 50 m/s ---------------------------------------------------------------- (a) Yes the kinetic energy conserved (b) Yes the momentum conserved (c) Apply conservation of momentum before and after collision m1*u1 = m2*v1 + m1*u2 v1 = (m1*u1 - m1*u2) / m2 = 1.875 m/s (d) Maximum height reached by the block is h = (v1)2 / 2g (where acceleration due to gravity g = 9.8 m/s2) = 0.18 m (e) Now from the figure, cos = (L - h) / L = (1.0 m - 0.18 m) / 1.0m = 0.82 Then we get = 34.85o (f) If the angle is 90o then maximum height reached by the block is h = L = 1.0m but the maximum height is h = (v1)2 / 2g (where acceleration due to gravity g = 9.8 m/s2) v1 = 2gh = 4.43 m/s From conservation of momemtum m1*u1 = m2*v1 + m1*u2 u1 = (m2*v1 + m1*u2 ) / m1 = 404.4 m/s v1 = 2gh = 4.43 m/s From conservation of momemtum m1*u1 = m2*v1 + m1*u2 u1 = (m2*v1 + m1*u2 ) / m1 = 404.4 m/s u1 = (m2*v1 + m1*u2 ) / m1 = 404.4 m/sRelated Questions
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