Momentum & Work problems: A 1 kg gun fires a 20 gram bullet that travels at 100 m/s...?
A 1 kg gun fires a 20 gram bullet that travels at 100 m/s. What will be the speed of recoil imparted to the gun, assuming that momentum is conserved perfectly (no heat generation)? A. 2 m/s B. 0.2 m/s C. 1.75 m/s D. 2.25 m/s E. None of the Above How much impulse did the gun mentioned above experience? A. 4 kg m/s B. 2 kg m/s C. 3 kg m/s D. 12 kg m/s E. None of the Above In a perfectly elastic collision, a 1 kg ball travelling at 2 m/s hits a 1 kg ball travelling at 1 m/s in the same direction. What is the total kinetic energy of this system? A. 0 kg m^2/s^2 B. 2.5 kg m^2/s^2 C. 12 kg m^2/s^2 D. 20 kg m^2/s^2 E. None of the Above What is the total momentum of the above system? A. 3 kg m/s B. 0 kg m/s C. 1 kg m/s D. 12 kg m/s Joey and Chandler are both sitting on one end of an 8 meter seesaw. Monica, who weighs 2000 N, comes along and sits on the other end, balancing the seesaw. If Joey leaves, Monica has to move 2.4 m closer to Chandler to balance the seesaw again. How much does Joey weigh? A. 750 N B. 800 N C. 900 N D. 1200 N E. None of the Above //My initial answer here is D but I'm not quite confident. Another friend, Rachel, who weighs 600 N, comes along and sits with Chandler. Monica has to move back to balance the seesaw. How far away will Monica be from Rachel and Chandler? A. 3 m B. 3.2 m C. 3.5 m D. 3.7 m E. None of the Above Brief explanations will do. :) I get the concepts of the problems but I can't seem to find out how to manipulate figures. Many thanks!
Physics - 2 Answers
People's Answers, Critics, Comments, Opinions :
Answer 1 :
1) By the law of momentum conservation:- =>m1u1+m2u2 = m1v1+m2v2 =>0 + 0 = 20 x 10^-3 x 100 + 1 x v2 =>v2 = -2 m/s [-ve just indicating the direction of gun recoil is opposite to the velocity of the bullet] =>A 2) By Impulse(I) = ∆P = m∆v = 1 x 2 = 2 kg-m/s =>B 3) KE(total) = 1/2m1u1^2 + 1/2m2u2^2 =>KE(t) = 1/2 x 1 x (2)^2 + 1/2 x 1 x (1)^2 =>KE(t) = 2.5 J =>B 4) P(total) = m1u1+m2u2 =>P(t) = 1 x 2 + 1 x 1 =>P(t) = 3 kg-m/s =>A 5)Let the weight of Chander is W1 Newton =>W1 x 4 = 2000 x (4-2.4) =>W1 = 800 Newton =>Joyes weight = 2000 - 800 = 1200 Newton =>D 6) Let Monika is now a meter from the center =>(800+600) x 4 = a x 2000 =>a = 2.8 m =>The distance of Monika from the two = 4 + 2.8 = 6.80 m =>E
Answer 2 :
1. Horizontal Momentum Conservation: m1v1 = m2v2 => 20/1000 * 100 = 1*v2 =>v2 = 2m/s is the recoil speed for the gun. Ans A 2.Impulse = Change in the momentum = Δp = p2-p1 = 2 kg m/s - 0 kg m/s = 2kg m/s. Ans B. 3. Here, both the particles have same mass, thus after the collision their speeds get interchanged. You can verify this from the formula. v1 = (m1-m2)u1+2m2u2/(m1+m2). v2 = (m2-m1)u2 + 2m1u1/(m1+m2). Thus the KE of the system = 1/2*m1*v1² + 1/2*m2*v2² = 1/2*1(v1²+v2²) = 0.5(2²+1²) = 2.5J. Ans B. 4. Total momentum = m1v1 + m2v2 = 1*2+1*1 = 3 kgm/s Ans A. 5. Let the force/weight of Joey and Chandler be J and C respectively. We've (J+C)4 = 2000*4 When Joey leaves, Monica moves 2.4m closer means the new distance is 1.6m 4C = 2000*1.6 =>C = 800 => J = 1200. Ans D You're correct. :) 6.Let the new distance be r for Monica from the axis of rotation. We've (600+800)*4 = 2000*r =>r = 1400*4/2000 = 2.8m Thus Monica is 6.8m(=4m + 2.8m) from Rachel and Chandler. Ans E Hope you get the solution.
A 1 kg gun fires a 20 gram bullet that travels at 100 m/s. What will be the speed of recoil imparted to the gun, assuming that momentum is conserved perfectly (no heat generation)? A. 2 m/s B. 0.2 m/s C. 1.75 m/s D. 2.25 m/s E. None of the Above How much impulse did the gun mentioned above experience? A. 4 kg m/s B. 2 kg m/s C. 3 kg m/s D. 12 kg m/s E. None of the Above In a perfectly elastic collision, a 1 kg ball travelling at 2 m/s hits a 1 kg ball travelling at 1 m/s in the same direction. What is the total kinetic energy of this system? A. 0 kg m^2/s^2 B. 2.5 kg m^2/s^2 C. 12 kg m^2/s^2 D. 20 kg m^2/s^2 E. None of the Above What is the total momentum of the above system? A. 3 kg m/s B. 0 kg m/s C. 1 kg m/s D. 12 kg m/s Joey and Chandler are both sitting on one end of an 8 meter seesaw. Monica, who weighs 2000 N, comes along and sits on the other end, balancing the seesaw. If Joey leaves, Monica has to move 2.4 m closer to Chandler to balance the seesaw again. How much does Joey weigh? A. 750 N B. 800 N C. 900 N D. 1200 N E. None of the Above //My initial answer here is D but I'm not quite confident. Another friend, Rachel, who weighs 600 N, comes along and sits with Chandler. Monica has to move back to balance the seesaw. How far away will Monica be from Rachel and Chandler? A. 3 m B. 3.2 m C. 3.5 m D. 3.7 m E. None of the Above Brief explanations will do. :) I get the concepts of the problems but I can't seem to find out how to manipulate figures. Many thanks!
Physics - 2 Answers
People's Answers, Critics, Comments, Opinions :
Answer 1 :
1) By the law of momentum conservation:- =>m1u1+m2u2 = m1v1+m2v2 =>0 + 0 = 20 x 10^-3 x 100 + 1 x v2 =>v2 = -2 m/s [-ve just indicating the direction of gun recoil is opposite to the velocity of the bullet] =>A 2) By Impulse(I) = ∆P = m∆v = 1 x 2 = 2 kg-m/s =>B 3) KE(total) = 1/2m1u1^2 + 1/2m2u2^2 =>KE(t) = 1/2 x 1 x (2)^2 + 1/2 x 1 x (1)^2 =>KE(t) = 2.5 J =>B 4) P(total) = m1u1+m2u2 =>P(t) = 1 x 2 + 1 x 1 =>P(t) = 3 kg-m/s =>A 5)Let the weight of Chander is W1 Newton =>W1 x 4 = 2000 x (4-2.4) =>W1 = 800 Newton =>Joyes weight = 2000 - 800 = 1200 Newton =>D 6) Let Monika is now a meter from the center =>(800+600) x 4 = a x 2000 =>a = 2.8 m =>The distance of Monika from the two = 4 + 2.8 = 6.80 m =>E
Answer 2 :
1. Horizontal Momentum Conservation: m1v1 = m2v2 => 20/1000 * 100 = 1*v2 =>v2 = 2m/s is the recoil speed for the gun. Ans A 2.Impulse = Change in the momentum = Δp = p2-p1 = 2 kg m/s - 0 kg m/s = 2kg m/s. Ans B. 3. Here, both the particles have same mass, thus after the collision their speeds get interchanged. You can verify this from the formula. v1 = (m1-m2)u1+2m2u2/(m1+m2). v2 = (m2-m1)u2 + 2m1u1/(m1+m2). Thus the KE of the system = 1/2*m1*v1² + 1/2*m2*v2² = 1/2*1(v1²+v2²) = 0.5(2²+1²) = 2.5J. Ans B. 4. Total momentum = m1v1 + m2v2 = 1*2+1*1 = 3 kgm/s Ans A. 5. Let the force/weight of Joey and Chandler be J and C respectively. We've (J+C)4 = 2000*4 When Joey leaves, Monica moves 2.4m closer means the new distance is 1.6m 4C = 2000*1.6 =>C = 800 => J = 1200. Ans D You're correct. :) 6.Let the new distance be r for Monica from the axis of rotation. We've (600+800)*4 = 2000*r =>r = 1400*4/2000 = 2.8m Thus Monica is 6.8m(=4m + 2.8m) from Rachel and Chandler. Ans E Hope you get the solution.
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